CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Gas Stoichiometry
Question Bank - Set 3
Liberty University
Question 1
Question
A gaseous compound containing hydrogen and nitrogen is decomposed with
heat to produce ammonia gas (NH3) and hydrogen gas (H2) according to the
following balanced chemical equation:
2NH2→2NH3+H2
If 3.50 grams of the gaseous compound decomposes, what is the volume of
the hydrogen gas produced at STP (standard temperature and pressure: 0
°
C
and 1 atm)? Assume all gases are ideal.
Solution
Step 1: Calculate the number of moles of the gaseous compound. Given mass
of the gaseous compound = 3.50 grams Molar mass of the gaseous compound
= 2(N) + 2(H) = 28 g/mol
Number of moles = 3.50 g
28 g/mol = 0.125 mol
Step 2: Use stoichiometry to determine the number of moles of hydrogen
gas produced. From the balanced chemical equation: 1 mol of the gaseous
compound produces 1 mol of hydrogen gas
Therefore, 0.125 mol of the gaseous compound produces 0.125 mol of hydro-
gen gas.
Step 3: Apply the ideal gas law to find the volume of hydrogen gas at STP.
Using the ideal gas law equation: P V =nRT
At STP, the conditions are 0
°
C (273 K) and 1 atm pressure.
Substitute: V=nRT
P=(0.125 mol)(0.0821 L·atm/mol·K)(273 K)
1 atm
Calculate: V= 2.96 L
Therefore, the volume of hydrogen gas produced at STP is 2.96 liters.
Question 2
Question
A sample of sodium hydrogen carbonate, NaHCO3, is decomposed by heating to
produce sodium carbonate, Na2CO3, carbon dioxide, and water vapor. If 50.0
grams of NaHCO3 is decomposed, what mass of sodium carbonate is produced?
Solution
Step 1: Write the balanced chemical equation for the decomposition of sodium
hydrogen carbonate:
2NaHCO3→Na2CO3+ CO2+ H2O
Step 2: Calculate the molar mass of each compound: - NaHCO3: 1 ×Na +
1×H+1×C+3×O = 23.0+1.0 + 12.0 + 48.0 = 84.0 g/mol - Na2CO3:
2×Na + 1 ×C+3×O=2×23.0 + 12.0 + 48.0 = 105.0 g/mol
Step 3: Calculate the number of moles of NaHCO3 using its molar mass and
the given mass:
moles of NaHCO3 = 50.0 g
84.0 g/mol = 0.595 mol
Step 4: Use the balanced chemical equation to find the molar ratio of
NaHCO3 to Na2CO3: From the equation, 2 moles of NaHCO3 produce 1 mole
of Na2CO3 (the coefficient ratio is 2:1).
Step 5: Calculate the mass of Na2CO3 produced by multiplying the moles
of NaHCO3 by the molar mass ratio:
mass of Na2CO3= 0.595 mol ×105.0 g
1 mol = 62.475 g
Therefore, when 50.0 grams of NaHCO3 is decomposed, 62.475 grams of
sodium carbonate is produced.
Question 3
Question
A gaseous compound containing only carbon and hydrogen is burned in oxygen
gas. If 2.00 grams of the compound produces 4.40 grams of carbon dioxide and
1.80 grams of water vapor, what is the molecular formula of the compound?
Given atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol
2
Solution
Step 1: Determine the number of moles of carbon dioxide and water vapor
produced.
Moles of CO2=4.40 g
44.01 g/mol = 0.100 mol
Moles of H2O = 1.80 g
18.02 g/mol = 0.100 mol
Step 2: Construct the chemical equation for the combustion of the com-
pound. Let the molecular formula of the compound be CxHy. The combustion
reaction will be:
CxHy+ (x+y)/4 O2→x CO2+ y/2 H2O
Step 3: Set up and solve a system of equations to find the values of x and y.
From the combustion reaction, we can write the following equations based on
the number of moles:
x= 0.1
y/2=0.1
12.01x+y/2×1.008 = 2.00
Solving the system of equations, we find that x = 2 and y = 4.
Step 4: Determine the molecular formula of the compound. Therefore, the
molecular formula of the gaseous compound is C2H4.
Question 4
Question
A gaseous compound contains only hydrogen and nitrogen. When 5.00 g of
the compound is decomposed, 1.42 g of nitrogen is produced. Determine the
empirical formula of the compound.
Given atomic masses: N = 14.01 g/mol, H = 1.008 g/mol.
Solution
Step 1: Find the moles of nitrogen produced.
Moles of nitrogen = Mass of nitrogen
Molar mass of nitrogen =1.42 g
14.01 g/mol
Moles of nitrogen ≈0.1015 mol
3
Step 2: Find the moles of hydrogen in the compound (since the remaining
mass is from hydrogen).
Moles of hydrogen = Mass of compound −Mass of nitrogen
Molar mass of hydrogen =5.00 g −1.42 g
1.008 g/mol
Moles of hydrogen ≈3.5417 mol
Step 3: Find the ratio of moles of hydrogen to nitrogen.
Moles of hydrogen
Moles of nitrogen =3.5417
0.1015 ≈34.90
Step 4: Determine the empirical formula by assuming 1 mole of nitrogen
and finding the nearest whole number ratio of hydrogen atoms. The empirical
formula is NH34.
Question 5
Question
When 15.0 g of ethane (C2H6) is burned in excess oxygen, how many grams of
water will be produced? Assume ethane reacts completely.
Solution
Step 1: Write the balanced chemical equation for the combustion of ethane:
C2H6+O2→CO2+H2O
Step 2: Calculate the molar mass of ethane (C2H6) and water (H2O) to use
in the stoichiometric calculation. Molar mass of C2H6: 2×molar mass of C +
6×molar mass of H = 2 ×12.01 g/mol + 6 ×1.01 g/mol = 30.07 g/mol
Molar mass of H2O: 2×molar mass of H + 1×molar mass of O = 2 ×
1.01 g/mol + 1 ×16.00 g/mol = 18.02 g/mol
Step 3: Calculate the number of moles of ethane in 15.0 g: Number of moles
=Mass
Molar mass =15.0 g
30.07 g/mol = 0.4989 mol
Step 4: Since ethane reacts completely, we can use the stoichiometry of the
balanced chemical equation to find the moles of water produced. From the bal-
anced chemical equation, 1 mole of ethane produces 3 moles of water. Therefore,
the number of moles of water produced = 0.4989 mol ×3=1.4967 mol
Step 5: Calculate the mass of water produced: Mass of water = Number of
moles ×Molar mass of H2OMass of water = 1.4967 mol×18.02 g/mol = 26.96 g
Therefore, when 15.0 g of ethane is burned in excess oxygen, 26.96 g of water
will be produced.
4
Question 6
Question
A gaseous compound containing only carbon and hydrogen is burned completely
in oxygen gas. If 2.00 L of the compound at STP produces 4.40 g of carbon
dioxide and 2.01 g of water, determine the empirical formula of the compound.
Solution
Step 1: Determine the moles of carbon dioxide produced. Given that 4.40 g of
carbon dioxide is produced, we can calculate the moles using the molar mass of
carbon dioxide (CO2). The molar mass of CO2is 44.01 g/mol.
Moles of CO2=4.40 g
44.01 g/mol = 0.1 mol
Step 2: Determine the moles of water produced. Given that 2.01 g of water
is produced, we can calculate the moles using the molar mass of water (H2O).
The molar mass of H2O is 18.02 g/mol.
Moles of H2O = 2.01 g
18.02 g/mol ≈0.1116 mol
Step 3: Calculate the moles of carbon and hydrogen atoms in the compound.
Since the compound contains only carbon and hydrogen, we can assume all of
the carbon in the compound ends up in the CO2and all of the hydrogen ends
up in the H2O. Thus, the moles of carbon and hydrogen can be calculated as
follows:
Moles of C = Moles of CO2= 0.1 mol
Moles of H = 2 ×Moles of H2O=2×0.1116 mol = 0.2232 mol
Step 4: Determine the empirical formula. To find the empirical formula, we
must determine the ratio of moles of carbon to moles of hydrogen. Dividing the
moles of each element by the smallest number of moles (0.1 mol), we get:
C:H
1:2.232
Thus, the empirical formula of the compound is CH2.
Question 7
Question
A 2.00 L container holds 2.00 g of nitrogen gas and 3.00 g of argon gas at a
certain temperature. If the mixture exerts a total pressure of 1.50 atm, what is
the partial pressure of each gas?
5
Solution
Step 1: Calculate the number of moles for each gas using the given masses and
molar masses. - The molar mass of nitrogen (N) is 28.02 g/mol and the molar
mass of argon (Ar) is 39.95 g/mol. - Let nN2be the number of moles of nitrogen
gas and nAr be the number of moles of argon gas.
nN2=2.00 g
28.02 g/mol = 0.0714 mol
nAr =3.00 g
39.95 g/mol = 0.0751 mol
Step 2: Calculate the total moles of gas in the container.
ntotal =nN2+nAr = 0.0714 mol + 0.0751 mol = 0.1465 mol
Step 3: Use the ideal gas law to find the partial pressure of each gas. - The
total pressure (Ptotal) is 1.50 atm. - The volume of the container (V) is 2.00
L. - The ideal gas constant (R) is 0.0821 L
·
atm/mol
·
K. - The temperature is
constant, so we can assume the temperature is in Kelvin.
Ptotal = (PN2+PAr) =⇒PN2+PAr = 1.50 atm
Since pressure is directly proportional to the number of moles of gas, we can
express the partial pressures in terms of the number of moles:
PN2=nN2
ntotal
×Ptotal =0.0714 mol
0.1465 mol ×1.50 atm = 0.732 atm
PAr =nAr
ntotal
×Ptotal =0.0751 mol
0.1465 mol ×1.50 atm = 0.768 atm
Therefore, the partial pressure of nitrogen gas is 0.732 atm and the partial
pressure of argon gas is 0.768 atm.
Question 8
Question
A 2.0 L container is filled with propane gas (C3H8) at a pressure of 3.0 atm and
a temperature of 25
°
C. If the propane reacts completely with excess oxygen gas
according to the following balanced equation:
C3H8+ 5O2→3CO2+ 4H2O
Calculate the volume of carbon dioxide (CO2) produced, assuming the re-
action goes to completion.
(Note: Assume all gases behave ideally.)
6
Solution
Step 1: Convert the temperature to Kelvin.
T(K) = T(C) + 273.15
T= 25C+ 273.15 = 298.15 K
Step 2: Use the ideal gas law to find the number of moles of propane.
P V =nRT
n=P V
RT =(3.0 atm ×2.0 L)
(0.0821 L ·atm/mol ·K×298.15 K)
n=6.0
24.51 = 0.244 mol
Step 3: Determine the limiting reactant. - Calculate the number of moles of
oxygen needed:
moles of O2= 5 ×moles of C3H8= 5 ×0.244 = 1.22 mol
Step 4: Calculate the volume of carbon dioxide produced. - From the bal-
anced chemical equation, we see that 1 mole of C3H8produces 3 moles of CO2.
moles of CO2= 3 ×moles of C3H8= 3 ×0.244 = 0.732 mol
- Use the ideal gas law to find the volume of CO2produced:
V=nRT
P=(0.732 mol ×0.0821 L ·atm/mol ·K×298.15 K)
3.0 atm
V=18.87
3.0= 6.29 L
Therefore, the volume of carbon dioxide (CO2) produced is 6.29 L.
Question 9
Question
A gaseous hydrocarbon fuel is composed of 85.0
Solution
Step 1: Calculate the moles of CO2produced. The molar mass of CO2is 44.01
g/mol. Given that the mass of CO2produced is 10.5 g, we have: Number of
moles of CO2=10.5 g
44.01 g/mol = 0.2386 mol.
Step 2: Calculate the moles of carbon in the hydrocarbon fuel. Assume
we have 100 g of the hydrocarbon fuel. Given that the fuel is composed of
7
85.0Mass of carbon = 100 g ×0.85 = 85 g. The molar mass of carbon is 12.01
g/mol. Number of moles of carbon = 85 g
12.01 g/mol = 7.076 mol.
Step 3: Calculate the moles of hydrogen in the hydrocarbon fuel. Given
that the fuel is composed of 15.0Mass of hydrogen = 100 g ×0.15 = 15 g.
The molar mass of hydrogen is 1.008 g/mol. Number of moles of hydrogen =
15 g
1.008 g/mol = 14.88 mol.
Step 4: Determine the empirical formula of the hydrocarbon fuel. The ratio
of moles of carbon to moles of hydrogen in the fuel is approximately 1:2 (to the
nearest whole number).
Therefore, the empirical formula of the hydrocarbon fuel is C1H2.
Step 5: Determine the molecular formula of the hydrocarbon fuel. The
molecular formula of the hydrocarbon fuel can be found using the molar mass
of the empirical formula and the molar mass of the actual compound.
The molar mass of C1H2is 1 ×12.01 + 2 ×1.008 = 14.03 g/mol.
The molar mass of the hydrocarbon fuel can be calculated as follows: Mo Grp molar mass
Empirical Formula molar mass =
3.45×Molar mass of hydrocarbon
10.5×Molar mass of CO2.
Solving for the molar mass of the hydrocarbon gives 42.09 g/mol.
Finally, the molecular formula of the hydrocarbon fuel is C3H6.
Question 10
Question
Given the reaction:
2 C3H8(g) + 7 O2(g)→6 CO2(g) + 8 H2O(g)
If 12.5 grams of propane (C3H8) reacts with excess oxygen gas, what mass of
water will be produced?
Solution
Step 1: Calculate the molar mass of C3H8. The molar mass of C3H8can be
calculated as:
Molar mass = 3 ×MC+ 8 ×MH
where MC= 12.01 g/mol and MH= 1.008 g/mol. Plugging in the values, we
get:
Molar mass = 3 ×12.01 + 8 ×1.008 = 44.10 g/mol
Step 2: Calculate the number of moles of propane. Use the formula:
Moles = Mass
Molar mass
Substitute the values:
Moles = 12.5
44.10 = 0.283 mol
8
Step 3: Use the mole ratio from the balanced equation to find the number of
moles of water produced. From the balanced equation, 2 moles of C3H8produce
8 moles of H2O. So, 0.283 moles of C3H8will produce:
0.283 mol ×8 mol H2O
2 mol C3H8
= 1.13 mol H2O
Step 4: Calculate the mass of water produced. Use the formula:
Mass = Moles ×Molar mass
Substitute the values:
Mass = 1.13 ×18.015 = 20.33 g
Therefore, 20.33 grams of water will be produced when 12.5 grams of propane
reacts with excess oxygen gas.
Question 11
Question
A sample of a gas with a volume of 4.50 L at a pressure of 1.20 atm and a
temperature of 25
°
C is reacted with excess O2to form SO2gas. If the volume
of SO2produced is 3.00 L at 1.20 atm and 25
°
C, what is the balanced chemical
equation for the reaction?
Solution
Step 1: Write the given information in terms of pressure, volume, and temper-
ature. Given: Initial gas: V1= 4.50 L, P1= 1.20 atm, T1= 25
°
C Final gas
(SO2): V2= 3.00 L, P2= 1.20 atm, T2= 25
°
C
Step 2: Use the ideal gas law to find the number of moles for each gas. For
the initial gas: n1=P1V1
RT1
For the final gas (SO2): n2=P2V2
RT2
Step 3: Since the initial gas reacted with excess O2, the moles of the initial
gas should be equal to the moles of the final gas (SO2). n1=n2
Step 4: Given that SO2is formed in the reaction, the balanced chemical
equation could be:
2A+ 2B→C
Where A and B represent the initial gases, and C represents the formed SO2
gas.
Therefore, the balanced chemical equation for the reaction is:
2A+ 2B→C
9
Question 12
Question
A gaseous hydrocarbon undergoes combustion according to the following bal-
anced chemical equation:
C6H14(g)+9O2(g)→6CO2(g)+7H2O(g)
If 15.0 grams of C6H14 react with excess oxygen, what volume of CO2gas will
be produced at STP (Standard Temperature and Pressure)? (1 mole of gas
occupies 22.4 L at STP)
Solution
Step 1: Calculate the moles of C6H14 in 15.0 grams: Given mass of C6H14: 15.0
grams Molar mass of C6H14: 6(12.01) + 14(1.01) = 86.18 g/mol
Moles of C6H14 =15.0 g
86.18 g/mol ≈0.174 mol
Step 2: Determine the moles of CO2produced using the stoichiometry of
the balanced equation: From the balanced equation, 1 mole of C6H14 produces
6 moles of CO2
Moles of CO2= 0.174 mol ×6 mol CO2
1 mol C6H14
= 1.044 mol
Step 3: Convert moles of CO2to volume at STP using the ideal gas law: At
STP, 1 mole of any gas occupies 22.4 L.
Volume of CO2= 1.044 mol ×22.4 L/mol = 23.462 L
Answer: The volume of CO2produced at STP is 23.462 L.
Question 13
Question
When 5.00 L of ethylene gas (C2H4) is reacted with excess oxygen gas, how many
liters of carbon dioxide gas (CO2) are produced at STP (Standard Temperature
and Pressure)? Assume all gases are ideal.
Solution
Step 1: Write the balanced chemical equation for the reaction between ethylene
and oxygen to produce carbon dioxide and water:
C2H4+ 3O2→2CO2+ 2H2O
10
Step 2: Determine the molar ratio of C2H4to CO2from the balanced equa-
tion: 1 mol of C2H4produces 2 mol of CO2
Step 3: Calculate the number of moles of C2H4using the ideal gas law at
STP:
P V =nRT
n=P V
RT =(1 atm)(5.00 L)
0.0821 L atm/mol K(273 K) = 0.196 mol
Step 4: Using the molar ratio from step 2, calculate the number of moles of
CO2produced:
0.196 mol ×2 mol CO2
1 mol C2H4
= 0.392 mol CO2
Step 5: Convert moles of CO2to volume at STP using the ideal gas law:
P V =nRT
V=nRT
P=(0.392 mol)(0.0821 L atm/mol K)(273 K)
1 atm = 8.17 L
Therefore, when 5.00 L of ethylene gas is reacted with excess oxygen gas,
8.17 L of carbon dioxide gas are produced at STP.
Question 14
Question
A gaseous mixture contains 16.0 g of nitrogen gas (N2) and 32.0 g of oxygen
gas (O2). If the gases react completely to form nitrogen monoxide gas (NO),
what is the mass of the product formed? Assume all gases are at the same
temperature and pressure.
Solution
Step 1: Determine the moles of each reactant using the given masses and molar
masses.
The molar mass of nitrogen gas (N2) is 28.02 g/mol, and the molar mass of
oxygen gas (O2) is 32.00 g/mol.
For nitrogen:
moles of N2=16.0 g
28.02 g/mol = 0.571 mol
For oxygen:
moles of O2=32.0 g
32.00 g/mol = 1.0 mol
Step 2: Determine the limiting reactant by looking at the mole ratio from
the balanced chemical equation.
11
The balanced chemical equation for the reaction is:
N2+ O2→2NO
From the equation, 1 mole of nitrogen gas reacts with 1 mole of oxygen gas
to produce 2 moles of nitrogen monoxide gas.
Since the mole ratio is 1:1, oxygen is the limiting reactant because it will be
completely consumed before all the nitrogen reacts.
Step 3: Calculate the mass of the product formed using the limiting reactant.
Using the mole ratio from the balanced chemical equation, we find that the
1.0 mol of oxygen will produce 2.0 mol of nitrogen monoxide.
The molar mass of nitrogen monoxide (NO) is 30.01 g/mol.
Mass of NO produced = 2.0 mol ×30.01 g/mol = 60.02 g
Therefore, the mass of the product formed when 16.0 g of nitrogen gas and
32.0 g of oxygen gas react completely is 60.02 g of nitrogen monoxide gas.
Question 15
Question
A mixture of 5.00 L of propane gas (C3H8) and 8.00 L of oxygen gas (O2)
are mixed in a container at 27
°
C and 1.00 atm pressure. The two gases react
according to the following balanced equation:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Assuming complete combustion of propane, what is the total volume of the
gaseous mixture at the same temperature and pressure after the reaction is
complete?
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
is:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Step 2: Calculate the moles of propane and oxygen present initially. Since
volume, temperature, and pressure are given, we can use the ideal gas law to
calculate the moles of each gas. The ideal gas law is given by:
P V =nRT
Rearranging to solve for moles, we have:
n=P V
RT
12
For propane:
nC3H8=(1.00 atm)(5.00 L)
0.0821 L ·atm/mol ·K·(300 K) = 0.205 moles
For oxygen:
nO2=(1.00 atm)(8.00 L)
0.0821 L ·atm/mol ·K·(300 K) = 0.328 moles
Step 3: Determine the limiting reactant. From the balanced equation, the
stoichiometric ratio of propane to oxygen is 1:5. Therefore, oxygen is the limiting
reactant since there is less oxygen available for the reaction.
Step 4: Calculate the volume of the gaseous mixture after the reaction is
complete. From the balanced equation, 1 mole of propane reacts with 5 moles of
oxygen to produce 3 moles of carbon dioxide and 4 moles of water. Since oxygen
is the limiting reactant with 0.328 moles, we can calculate the moles of carbon
dioxide and water produced: - Moles of CO2: 0.328 moles ×3
5= 0.197 moles -
Moles of H2O: 0.328 moles ×4
5= 0.262 moles
Now we can use the ideal gas law to find the volume of the gaseous mixture
after the reaction:
V=nRT
P
V=(0.197 moles + 0.262 moles)(0.0821 L ·atm/mol ·K·(300 K))
1.00 atm
V= 12.1 L
Therefore, the total volume of the gaseous mixture after the reaction is
complete is 12.1 L.
Question 16
Question
When 32.0 g of magnesium metal (Mg) react with excess hydrochloric acid
(HCl), how many liters of hydrogen gas (H2) at STP are produced?
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and hydrochloric acid:
Mg + 2HCl →MgCl2+H2
Step 2: Determine the number of moles of magnesium used: Given mass of
magnesium (Mg): 32.0 g Molar mass of Mg: 24.305 g/mol
Number of moles of Mg =32.0 g
24.305 g/mol ≈1.32 mol
13
Step 3: Use the mole ratio from the balanced chemical equation to find the
number of moles of H2: From the balanced equation, 1 mol of Mg produces 1
mol of H2.
Number of moles of H2= 1.32 mol
Step 4: Calculate the volume of hydrogen gas at STP (Standard Tempera-
ture and Pressure): 1 mol of any gas at STP occupies 22.4 L.
Volume of H2gas at STP = 1.32 mol ×22.4 L/mol = 29.57 L
Step 5: Answer: The reaction of 32.0 g of magnesium with excess hy-
drochloric acid will produce 29.57 L of hydrogen gas at STP.
Question 17
Question
A 2.00 L container holds 0.400 mol of nitrogen gas and 0.800 mol of hydrogen
gas at 500 K. If the gases react to form ammonia gas (NH3) according to the
balanced equation:
N2(g)+3H2(g)→2NH3(g)
What is the limiting reactant? How many moles of ammonia gas can be
produced?
Solution
Step 1: Write the balanced chemical equation
N2(g)+3H2(g)→2NH3(g)
Step 2: Calculate the number of moles of ammonia that can be produced
from nitrogen gas For every mole of nitrogen gas, 2 moles of ammonia are
produced. Therefore, the maximum number of moles of ammonia that can be
produced from nitrogen gas is:
0.400 mol ×2 mol NH3
1 mol N2
= 0.800 mol NH3
Step 3: Calculate the number of moles of ammonia that can be produced
from hydrogen gas For every 3 moles of hydrogen gas, 2 moles of ammonia are
produced. Therefore, the maximum number of moles of ammonia that can be
produced from hydrogen gas is:
0.800 mol ×2 mol NH3
3 mol H2
= 0.533 mol NH3
Step 4: Determine the limiting reactant Since nitrogen gas produces a greater
amount of ammonia (0.800 mol) compared to hydrogen gas (0.533 mol), hydro-
gen gas is the limiting reactant.
14
Step 5: Calculate the number of moles of ammonia gas that can be produced
The limiting reactant is hydrogen gas, which produces 0.533 mol of ammonia.
Thus, 0.533 mol of ammonia gas can be produced.
Question 18
Question
A reaction between acetylene (C2H2) and oxygen produces carbon dioxide and
water. If 10.0 L of acetylene gas at STP is reacted with an excess of oxygen,
what volume of carbon dioxide gas is produced at STP?
Solution
Step 1: Write the balanced chemical equation for the reaction between acetylene
and oxygen:
C2H2+ O2→CO2+ H2O
Step 2: Determine the stoichiometry of the reaction. From the balanced
chemical equation, we see that 1 mole of acetylene produces 1 mole of carbon
dioxide gas.
Step 3: Calculate the number of moles of acetylene that react:
Moles of C2H2=Volume of C2H2
Molar volume at STP =10.0 L
22.4 L/mol = 0.4464 moles
Step 4: Determine the volume of carbon dioxide gas produced at STP using
the Ideal Gas Law:
PV = nRT
VCO2=nCO2RT
P=(0.4464 moles)(0.0821 atm ·L/mol ·K)(273 K)
1 atm
VCO2= 10.4 L
Therefore, 10.4 L of carbon dioxide gas is produced at STP when 10.0 L of
acetylene gas is reacted with excess oxygen.
Question 19
Question
A reaction between solid iron and gaseous sulfur produces solid iron(II) sulfide
according to the following balanced chemical equation:
Fe(s)+S2(g)→FeS(s)
If 5.00 g of iron reacts with an excess of sulfur and produces 8.72 g of iron(II)
sulfide, determine the theoretical and percent yields of the reaction.
15
Solution
Step 1: Calculate the molar mass of iron(II) sulfide. The molar mass of FeS is:
Molar mass of FeS = Mass of Fe+Mass of S = 55.85 g/mol+32.07 g/mol = 87.92 g/mol
Step 2: Determine the number of moles of iron used. Given:
Mass of Fe = 5.00 g
Molar mass of Fe = 55.85 g/mol
Calculate the number of moles of iron:
Moles of Fe = Mass of Fe
Molar mass of Fe =5.00 g
55.85 g/mol = 0.0894 mol
Step 3: Determine the limiting reactant. From the balanced chemical equa-
tion, the stoichiometry of the reaction is 1:1 between Fe and FeS. Therefore, the
limiting reactant is Fe.
Step 4: Calculate the theoretical yield of iron(II) sulfide. Using the mole
ratio from the balanced chemical equation, the theoretical yield can be calcu-
lated:
Moles of FeS produced = Moles of Fe = 0.0894 mol
Calculate the mass of FeS produced:
Mass of FeS produced = Moles of FeS×Molar mass of FeS = 0.0894 mol×87.92 g/mol = 7.86 g
Step 5: Calculate the percent yield of the reaction. Given:
Actual yield = 8.72 g
Theoretical yield = 7.86 g
Calculate the percent yield:
Percent yield = Actual yield
Theoretical yield ×100% = 8.72 g
7.86 g ×100% = 110.89%
Therefore, the theoretical yield of iron(II) sulfide is 7.86 g and the percent
yield of the reaction is 110.89
Question 20
Question
A chemical reaction takes place in a sealed container. The reaction involves the
combustion of propane gas (C3H8) with oxygen gas to produce carbon dioxide
gas and water vapor. If 4.50 L of propane gas reacts with excess oxygen gas at
STP (standard temperature and pressure), what volume of carbon dioxide gas
is produced? Assume all gases are ideal.
16
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the combustion of propane is:
C3H8+ 5O2→3CO2+ 4H2O
Step 2: Determine the moles of propane gas. Using the ideal gas law equation
P V =nRT , we can find the moles of propane gas. Given: Volume (V) = 4.50
L Temperature (T) = 273 K (STP) Pressure (P) = 1 atm Gas constant (R) =
0.0821 L atm/(mol K)
From the ideal gas law:
n=P V
RT =(1 atm)(4.50 L)
(0.0821 L atm/(mol K))(273 K)
Calculating ngives:
n≈0.184 moles
Step 3: Use the stoichiometry of the reaction to find the volume of carbon
dioxide gas produced. From the balanced equation, we know that 1 mole of
propane produces 3 moles of carbon dioxide gas. Therefore, 0.184 moles of
propane produces:
(0.184 moles propane) ×3 moles CO2
1 mole C3H8= 0.552 moles CO2
Now, using the ideal gas law again with the new number of moles, we can
calculate the volume of carbon dioxide gas produced:
V=nRT
P=(0.552 moles)(0.0821 L atm/(mol K))(273 K)
1 atm
Calculating Vgives:
V≈12.8 L
Therefore, approximately 12.8 L of carbon dioxide gas is produced during
the reaction.
Question 21
Question
Calculate the volume of carbon monoxide gas (CO) produced at 273 K and 1
atm when 10.0 g of carbon (C) reacts with excess carbon dioxide gas (CO2)
according to the following balanced chemical equation:
C(s) + CO2(g)→2CO(g)
17
Solution
Step 1: Calculate the number of moles of carbon (C) used.
Molar mass of C= 12.01 g/mol
Moles of C=10.0 g
12.01 g/mol
Moles of C= 0.832 mol
Step 2: Use stoichiometry to find the number of moles of carbon monoxide
gas (CO) produced.
From the balanced equation: 1 mol C→2 mol CO
0.832 mol C→0.832 mol ×2
0.832 mol C→1.664 mol CO
Step 3: Convert moles of carbon monoxide gas to volume at 273 K and 1
atm using the ideal gas law equation P V =nRT .
P= 1 atm
V=?
n= 1.664 mol
R= 0.0821 L ·atm/mol ·K
T= 273 K
Step 4: Solve for the volume of carbon monoxide gas (CO).
V=nRT
P
V=(1.664 mol)(0.0821 L ·atm/mol ·K)(273 K)
1 atm
V= 37.1 L
Therefore, 37.1 L of carbon monoxide gas (CO) is produced.
Question 22
Question
A gaseous compound contains only sulfur and oxygen. When 6.25 g of the
compound is decomposed, 2.35 g of sulfur and 3.90 g of oxygen are produced.
What is the formula of the compound?
(Note: The molar masses are as follows: S = 32.07 g/mol, O = 16.00 g/mol)
18
Solution
Step 1: Calculate the number of moles of sulfur and oxygen produced.
Moles of Sulfur = 2.35 g
32.07 g/mol = 0.0732 mol
Moles of Oxygen = 3.90 g
16.00 g/mol = 0.2438 mol
Step 2: Determine the mole ratio of sulfur to oxygen.
Mole ratio (Sulfur : Oxygen) = 0.0732 mol
0.2438 mol = 0.3:1
Step 3: Find the empirical formula of the compound. Since the formula
contains only sulfur and oxygen, the empirical formula of the compound is SO3.
Therefore, the formula of the compound is sulfur trioxide (SO3).
Question 23
Question
A sample of butane gas (C4H10) is burned in excess oxygen gas (O2) to produce
carbon dioxide gas (CO2) and water vapor (H2O). If 38.0 g of butane reacts
with 280.0 g of oxygen gas, what is the limiting reactant and how many grams
of water vapor are produced?
Solution
Step 1: Write the balanced chemical equation for the reaction:
2C4H10 + 13O2→8CO2+ 10H2O
Step 2: Calculate the molar mass of each substance:
Molar mass of C4H10 = 4(12.01) + 10(1.01) = 58.12 g/mol
Molar mass of O2= 2(16.00) = 32.00 g/mol
Molar mass of CO2= 12.01 + 2(16.00) = 44.01 g/mol
Molar mass of H2O= 2(1.01) + 16.00 = 18.02 g/mol
Step 3: Calculate the number of moles of each reactant:
Moles of C4H10 =38.0 g
58.12 g/mol = 0.654 mol
Moles of O2=280.0 g
32.00 g/mol = 8.75 mol
19
Step 4: Determine the limiting reactant by comparing the mole ratios from
the balanced chemical equation:
Moles of C4H10∇ · 2=0.327 mol
Moles of O2∇ · 13 = 0.673 mol
Since there is more of the butane, it is in excess and oxygen is the limiting
reactant.
Step 5: Calculate the theoretical yield of water vapor:
Moles of water vapor = 8.75 mol ×10
13 = 6.73 mol
Mass of water vapor = 6.73 mol ×18.02 g/mol = 121.0 g
Therefore, the limiting reactant is oxygen gas and 121.0 grams of water vapor
are produced.
Question 24
Question
A sample of a gaseous hydrocarbon undergoes combustion in oxygen gas, pro-
ducing carbon dioxide and water vapor. If 6.40 L of carbon dioxide are produced
at STP, how many moles of the hydrocarbon were in the original sample?
Solution
Step 1: Write the balanced chemical equation for the combustion of the hydro-
carbon. The general formula for a hydrocarbon is CxHy. The balanced chemical
equation for the combustion of a hydrocarbon is:
CxHy+ (x+y/4)O2→xCO2+ (y/2)H2O
Step 2: Calculate the volume of carbon dioxide produced. At STP (Standard
Temperature and Pressure), 1 mole of any gas occupies 22.4 L. The volume of
carbon dioxide produced is 6.40 L. We can use this information to find the
number of moles of carbon dioxide produced:
Moles of CO2=V olume
22.4=6.40
22.4= 0.285 moles
Step 3: Determine the moles of the hydrocarbon. From the balanced chem-
ical equation, we see that the mole ratio of CxHyto CO2is 1:1. Hence, the
moles of the hydrocarbon are also 0.285 moles.
Step 4: Answer Therefore, there were 0.285 moles of the hydrocarbon in the
original sample.
20
Question 25
Question
A 2.50 L container at 27.0
°
C contains 3.00 moles of nitrogen gas and 2.00 moles
of hydrogen gas. The hydrogen gas reacts with the nitrogen gas to form ammonia
gas and no hydrogen is left over. After the reaction, the temperature of the
mixture is 37.0
°
C. What is the pressure in the container after the reaction?
Assume all gases are ideal.
Solution
Step 1: Calculate the initial pressure of the gases using the ideal gas law:
P V =nRT
For nitrogen gas:
PN2=nN2RT
V=(3.00 mol)(0.0821 L ·atm/mol ·K)(27.0 + 273.15 K)
2.50 L
PN2≈6.35 atm
For hydrogen gas:
PH2=nH2RT
V=(2.00 mol)(0.0821 L ·atm/mol ·K)(27.0 + 273.15 K)
2.50 L
PH2≈4.24 atm
Step 2: Write the balanced chemical equation:
3 H2(g)+N2(g)→2 NH3(g)
Step 3: Determine the limiting reactant. Since the hydrogen gas is com-
pletely consumed and nitrogen gas is in excess, nitrogen gas is the limiting
reactant.
Step 4: Calculate the moles of ammonia produced using the stoichiometry of
the reaction: Moles of ammonia produced = 3.00 mol N2
1×2 mol NH3
1 mol N2= 6.00 moles
NH3
Step 5: Calculate the final pressure of the mixture using the ideal gas law:
PNH3=nNH3RT
V=(6.00 mol)(0.0821 L ·atm/mol ·K)(37.0 + 273.15 K)
2.50 L
PNH3≈13.31 atm
Thus, the pressure in the container after the reaction is approximately 13.31
atm.
21
Question 26
Question
A gas mixture contains 6.0 moles of oxygen gas and 4.0 moles of hydrogen gas.
If the gases react according to the equation 2H2(g) + O2(g) →2H2O(g), what
is the limiting reagent?
Solution
Step 1: Write and balance the chemical equation. The balanced chemical equa-
tion is:
2H2(g) + O2(g) →2H2O(g)
This equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen
gas to produce 2 moles of water vapor.
Step 2: Determine the mole ratio of reactants. From the balanced equation,
the mole ratio of H2to O2is 2:1. This means that for every 2 moles of H2, 1
mole of O2is needed to react completely.
Step 3: Calculate the moles of O2needed for the reaction with the available
moles of reactants. For 6.0 moles of oxygen gas, we need (6.0 moles O2)×
(2 moles H2/1 mole O2) = 12.0 moles H2.
Step 4: Compare the moles of O2needed with the moles of H2available.
Since there are only 4.0 moles of hydrogen gas available, which is less than the
12.0 moles required, hydrogen gas is the limiting reagent.
Therefore, hydrogen gas is the limiting reagent in this reaction.
Question 27
Question
A gaseous compound consists of nitrogen and oxygen. If 5.00 L of the compound
at STP decomposes to form 2.00 L of nitrogen gas and the remaining volume is
oxygen gas, determine the empirical formula of the compound. Assume that all
gases are ideal.
Solution
Step 1: Identify the gases involved in the reaction. Let’s denote the compound
as XnOm, where X represents an element and n and m are integers.
Step 2: Determine the moles of nitrogen gas produced. From the ideal gas
law, we can determine the number of moles of nitrogen gas:
molN2=P V
RT =(1.00 atm ×2.00 L)
(0.0821 L ·atm/mol ·K×273 K) = 0.0917 mol
22
Step 3: Determine the moles of oxygen gas produced. Since we know the
total volume is 5.00 L and that 2.00 L is nitrogen gas, the remaining volume
must be oxygen gas:
VolO2= 5.00 L −2.00 L = 3.00 L
Now we can determine the moles of oxygen gas:
molO2=P V
RT =(1.00 atm ×3.00 L)
(0.0821 L ·atm/mol ·K×273 K) = 0.137 mol
Step 4: Determine the ratio of moles of nitrogen and oxygen. Dividing the
moles of oxygen gas by the moles of nitrogen gas gives us the empirical formula:
molO2
molN2
=0.137
0.0917 = 1.49 ≈3
2
Therefore, the empirical formula of the compound is N2O3.
Question 28
Question
A reaction occurs between nitrogen dioxide gas (N O2) and water vapor (H2O) to
produce nitric acid (HNO3) and nitrogen monoxide gas (NO). If 5.00 L of NO2
react with an excess of water vapor at standard temperature and pressure, how
many liters of nitrogen monoxide gas will be produced based on the following
balanced chemical equation?
3NO2(g) + H2O(g)→2HNO3(aq) + N O(g)
Solution
Step 1: Write down the balanced chemical equation for the reaction.
3NO2(g) + H2O(g)→2HNO3(aq) + N O(g)
Step 2: Determine the stoichiometry of the reaction. From the balanced
equation, we see that for every 3 moles of N O2that react, 1 mole of N O is
produced.
Step 3: Convert the volume of NO2to moles using the ideal gas law. Given
that 1 mole of any gas occupies 22.4 L at standard temperature and pressure
(STP), we have:
5.00L×1 mole
22.4L= 0.2232 moles of NO2
Step 4: Determine the volume of NO produced using the mole ratio from
the balanced equation. Since 3 moles of NO2produce 1 mole of NO, we have:
0.2232 moles NO2×1 mole NO
3 moles NO2
= 0.0744 moles NO
23
Step 5: Convert moles of NO to volume at STP. Using the ideal gas law, we
find:
0.0744 moles NO ×22.4L= 1.67 L NO
Answer: The reaction will produce 1.67 liters of nitrogen monoxide gas.
Question 29
Question
A gaseous compound contains only carbon and hydrogen. Combustion of a
0.750 g sample of the compound produced 2.274 g of carbon dioxide and 1.165
g of water. Determine the empirical formula of the compound.
(Assume that all measurements are accurate, and the combustion is com-
plete.)
Solution
Step 1: Calculate the moles of carbon dioxide produced. Given: Mass of carbon
dioxide produced = 2.274 g Molar mass of carbon dioxide (CO2) = 44.01 g/mol
Moles of CO2=Mass of CO2
Molar mass of CO2
Moles of CO2=2.274 g
44.01 g/mol
Moles of CO2≈0.0517 mol
Step 2: Calculate the moles of water produced. Given: Mass of water pro-
duced = 1.165 g Molar mass of water (H2O) = 18.02 g/mol
Moles of H2O = Mass of H2O
Molar mass of H2O
Moles of H2O = 1.165 g
18.02 g/mol
Moles of H2O≈0.0647 mol
Step 3: Determine the moles of carbon and hydrogen in the compound. From
the balanced chemical equation for the combustion reaction of the compound,
we know that 1 mole of C will produce 1 mole of CO2and that 1 mole of H2O
will produce 1 mole of H2O.
Therefore, the ratio of moles of carbon to moles of hydrogen in the compound
is:
Moles of C : Moles of H = 0.0517 mol
0.0647 mol
Simplify the ratio:
Moles of C : Moles of H ≈0.799 : 1
24
Step 4: Determine the empirical formula of the compound. Since the ratio of
moles of carbon to moles of hydrogen is approximately 0.799 : 1, the empirical
formula of the compound is C4H5.
Question 30
Question
A mixture of ethane (C2H6) and oxygen gas (O2) reacts according to the fol-
lowing balanced chemical equation:
C2H6(g) + 7
2O2(g)→2CO2(g) + 3H2O(g)
If 3.00 grams of ethane are reacted with excess oxygen, what mass of water
would be produced?
Solution
Step 1: Calculate the molar mass of ethane (C2H6):
Molar mass of C2H6= 2 ×molar mass of C + 6 ×molar mass of H
= 2 ×12.01 g/mol + 6 ×1.01 g/mol = 30.07 g/mol
Step 2: Calculate the number of moles of ethane used:
Moles of C2H6=Mass
Molar mass =3.00 g
30.07 g/mol = 0.0998 mol
Step 3: Determine the limiting reactant: In this case, ethane is the limiting
reactant because it is given that there is excess oxygen.
Step 4: Calculate the theoretical yield of water produced using the mole
ratio from the balanced chemical equation:
Moles of H2O = Moles of C2H6×3 mol H2O
1 mol C2H6
= 0.0998 mol×3 mol H2O
1 mol C2H6
= 0.299 mol
Step 5: Calculate the mass of water produced:
Mass of H2O = Moles×Molar mass of H2O = 0.299 mol×(2×1.01+16.00) g/mol = 9.02 g
Therefore, 9.02 grams of water would be produced when 3.00 grams of ethane
are reacted with excess oxygen.
25
Question 31
Question
A gaseous compound containing only nitrogen and oxygen is decomposed by
passing an electric current through it. If 37.46 mL of nitrogen gas at 25
°
C and
0.987 atm is produced and collected over water at 25
°
C, what volume of oxygen
gas at 25
°
C and 0.987 atm is collected, assuming the vapor pressure of water
at 25
°
C is 23.8 mmHg?
Solution
Step 1: Calculate the pressure of the gas collected over water. Given that the
vapor pressure of water at 25
°
C is 23.8 mmHg, we need to convert this to atm
to use it in calculations. 1 atm = 760 mmHg, therefore 23.8 mmHg = 23.8/760
atm = 0.03132 atm.
The pressure of the gas collected over water is 0.987 atm (given) −0.03132
atm = 0.95568 atm.
Step 2: Calculate the number of moles of nitrogen gas produced. Using the
ideal gas law P V =nRT , we can calculate the number of moles of nitrogen gas
produced. n1=P V
RT , where Pis the pressure, Vis the volume, Ris the ideal
gas constant, and Tis the temperature in Kelvin. R= 0.0821 L atm/mol K,
T= 25 + 273.15 K.
n1=(0.987)(0.03746/1000)
(0.0821)(298.15)
n10.00000177 mol.
Step 3: Determine the mole ratio between nitrogen and oxygen. The bal-
anced chemical equation for the decomposition of the compound will help us
establish the mole ratio between the gases.
Step 4: Calculate the volume of oxygen gas collected. Since oxygen gas is
twice the number of moles of nitrogen gas, the volume of oxygen gas collected will
be twice that of nitrogen gas. 0.00000177 mol of oxygen gas is produced. Using
the ideal gas law P V =nRT :V2=n2RT
PV2=(2 ×0.00000177) ×(0.0821) ×(298.15)
0.95568
V2=0.00000354 ×24.5585
0.95568
V20.0906 L ≈90.6 mL.
Therefore, approximately 90.6 mL of oxygen gas at 25
°
C and 0.987 atm is
collected.
26
Question 32
Question
A certain amount of propane gas (C3H8) is burned in air. The reaction is as
follows:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
If 25.0 g of propane are burned, what volume of CO2gas is produced at a
temperature of 298 K and a pressure of 1.00 atm?
(Molar mass of C3H8= 44.1 g/mol, CO2= 44.0 g/mol, H2O= 18.0 g/mol,
1 mol CO2at STP occupies 22.4 L)
Solution
Step 1: Convert the mass of propane (C3H8) to moles using its molar mass.
Moles of C3H8=25.0 g
44.1 g/mol
= 0.567 mol
Step 2: Use the stoichiometry of the reaction to find the moles of CO2
that will be produced. Since the balanced equation shows that 1 mole of C3H8
produces 3 moles of CO2, we have:
Moles of CO2= 0.567 mol ×3 mol CO2
1 mol C3H8
= 1.70 mol
Step 3: Calculate the volume of CO2gas produced using the ideal gas law.
The ideal gas law is:
P V =nRT
Given: - n= 1.70 mol, - T= 298 K, - P= 1.00 atm, - R= 0.0821 atm ·
L/mol ·K.
First, convert the temperature to kelvin:
T= 298 K
Now, plug in the values and solve for V:
V=nRT
P
=(1.70 mol)(0.0821 atm ·L/mol ·K)(298 K)
1.00 atm
= 41.7 L
Therefore, the volume of CO2gas produced is 41.7 L.
27
Question 33
Question
When sodium azide (NaN3) is heated, it decomposes into sodium metal and
nitrogen gas. If 10.0 grams of sodium azide is heated, what mass of nitrogen
gas is produced? (Assume all reactants are used up in the reaction.)
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the decomposition of sodium azide is:
2NaN3→2Na + 3N2
Step 2: Calculate the molar mass of NaN3. The molar mass of NaN3:
M(Na) = 22.99 g/mol
M(N) = 14.01 g/mol
M(N3)=3×M(N) = 3 ×14.01 = 42.03 g/mol
M(NaN3) = M(Na) + M(N3) = 22.99 + 42.03 = 65.02 g/mol
Step 3: Calculate the number of moles of NaN3in 10.0 grams.
Number of moles = Mass
Molar mass =10.0 g
65.02 g/mol ≈0.153 mol
Step 4: Use the mole ratio from the balanced equation to calculate the
number of moles of N2produced. From the balanced equation, the mole ratio
of NaN3to N2is 2:3.
Moles of N2= 0.153 mol ×3 mol N2
2 mol NaN3
= 0.229 mol
Step 5: Calculate the mass of N2produced. Using the molar mass of nitrogen
gas (N2):
M(N2)=2×M(N) = 2 ×14.01 = 28.02 g/mol
Mass of N2= Moles ×Molar mass = 0.229 mol ×28.02 g/mol ≈6.42 g
Therefore, when 10.0 grams of sodium azide is heated, approximately 6.42
grams of nitrogen gas is produced.
Question 34
Question
A gaseous compound contains only carbon and hydrogen. When 0.500 g of the
compound is completely burned in oxygen, 1.18 g of carbon dioxide and 0.482
g of water are produced. Determine the empirical formula of the compound.
28
Solution
Step 1: Determine the number of moles of carbon dioxide and water produced.
moles of CO2=1.18 g
44.01 g/mol
= 0.0268 mol
moles of H2O=0.482 g
18.015 g/mol
= 0.0268 mol
Step 2: Determine the moles of carbon and hydrogen in the original com-
pound using the mole ratio in the balanced chemical equation.
moles of C= moles of CO2
= 0.0268 mol
moles of H= 2 ×moles of H2O
= 2 ×0.0268 mol
= 0.0536 mol
Step 3: Determine the empirical formula of the compound.
Molar ratio of C:H=0.0268 mol
0.0536 mol
= 1 : 2
Therefore, the empirical formula of the compound is CH2.
Question 35
Question
If 4.50 L of carbon monoxide gas reacts with 3.25 L of oxygen gas at the same
conditions of temperature and pressure, what volume of carbon dioxide gas
would be produced according to the following balanced chemical equation?
2CO(g)+O2(g)→2CO2(g)
Solution
Step 1: Write down the balanced chemical equation.
CO(g) + 1
2O2(g)→CO2(g)
29
Question 2
Question
A sample of sodium hydrogen carbonate, NaHCO3, is decomposed by heating to
produce sodium carbonate, Na2CO3, carbon dioxide, and water vapor. If 50.0
grams of NaHCO3 is decomposed, what mass of sodium carbonate is produced?
Solution
Step 1: Write the balanced chemical equation for the decomposition of sodium
hydrogen carbonate:
2NaHCO3→Na2CO3+ CO2+ H2O
Step 2: Calculate the molar mass of each compound: - NaHCO3: 1 ×Na +
1×H+1×C+3×O = 23.0+1.0 + 12.0 + 48.0 = 84.0 g/mol - Na2CO3:
2×Na + 1 ×C+3×O=2×23.0 + 12.0 + 48.0 = 105.0 g/mol
Step 3: Calculate the number of moles of NaHCO3 using its molar mass and
the given mass:
moles of NaHCO3 = 50.0 g
84.0 g/mol = 0.595 mol
Step 4: Use the balanced chemical equation to find the molar ratio of
NaHCO3 to Na2CO3: From the equation, 2 moles of NaHCO3 produce 1 mole
of Na2CO3 (the coefficient ratio is 2:1).
Step 5: Calculate the mass of Na2CO3 produced by multiplying the moles
of NaHCO3 by the molar mass ratio:
mass of Na2CO3= 0.595 mol ×105.0 g
1 mol = 62.475 g
Therefore, when 50.0 grams of NaHCO3 is decomposed, 62.475 grams of
sodium carbonate is produced.
Question 3
Question
A gaseous compound containing only carbon and hydrogen is burned in oxygen
gas. If 2.00 grams of the compound produces 4.40 grams of carbon dioxide and
1.80 grams of water vapor, what is the molecular formula of the compound?
Given atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol
2
Solution
Step 1: Determine the number of moles of carbon dioxide and water vapor
produced.
Moles of CO2=4.40 g
44.01 g/mol = 0.100 mol
Moles of H2O = 1.80 g
18.02 g/mol = 0.100 mol
Step 2: Construct the chemical equation for the combustion of the com-
pound. Let the molecular formula of the compound be CxHy. The combustion
reaction will be:
CxHy+ (x+y)/4 O2→x CO2+ y/2 H2O
Step 3: Set up and solve a system of equations to find the values of x and y.
From the combustion reaction, we can write the following equations based on
the number of moles:
x= 0.1
y/2=0.1
12.01x+y/2×1.008 = 2.00
Solving the system of equations, we find that x = 2 and y = 4.
Step 4: Determine the molecular formula of the compound. Therefore, the
molecular formula of the gaseous compound is C2H4.
Question 4
Question
A gaseous compound contains only hydrogen and nitrogen. When 5.00 g of
the compound is decomposed, 1.42 g of nitrogen is produced. Determine the
empirical formula of the compound.
Given atomic masses: N = 14.01 g/mol, H = 1.008 g/mol.
Solution
Step 1: Find the moles of nitrogen produced.
Moles of nitrogen = Mass of nitrogen
Molar mass of nitrogen =1.42 g
14.01 g/mol
Moles of nitrogen ≈0.1015 mol
3
Step 2: Find the moles of hydrogen in the compound (since the remaining
mass is from hydrogen).
Moles of hydrogen = Mass of compound −Mass of nitrogen
Molar mass of hydrogen =5.00 g −1.42 g
1.008 g/mol
Moles of hydrogen ≈3.5417 mol
Step 3: Find the ratio of moles of hydrogen to nitrogen.
Moles of hydrogen
Moles of nitrogen =3.5417
0.1015 ≈34.90
Step 4: Determine the empirical formula by assuming 1 mole of nitrogen
and finding the nearest whole number ratio of hydrogen atoms. The empirical
formula is NH34.
Question 5
Question
When 15.0 g of ethane (C2H6) is burned in excess oxygen, how many grams of
water will be produced? Assume ethane reacts completely.
Solution
Step 1: Write the balanced chemical equation for the combustion of ethane:
C2H6+O2→CO2+H2O
Step 2: Calculate the molar mass of ethane (C2H6) and water (H2O) to use
in the stoichiometric calculation. Molar mass of C2H6: 2×molar mass of C +
6×molar mass of H = 2 ×12.01 g/mol + 6 ×1.01 g/mol = 30.07 g/mol
Molar mass of H2O: 2×molar mass of H + 1×molar mass of O = 2 ×
1.01 g/mol + 1 ×16.00 g/mol = 18.02 g/mol
Step 3: Calculate the number of moles of ethane in 15.0 g: Number of moles
=Mass
Molar mass =15.0 g
30.07 g/mol = 0.4989 mol
Step 4: Since ethane reacts completely, we can use the stoichiometry of the
balanced chemical equation to find the moles of water produced. From the bal-
anced chemical equation, 1 mole of ethane produces 3 moles of water. Therefore,
the number of moles of water produced = 0.4989 mol ×3=1.4967 mol
Step 5: Calculate the mass of water produced: Mass of water = Number of
moles ×Molar mass of H2OMass of water = 1.4967 mol×18.02 g/mol = 26.96 g
Therefore, when 15.0 g of ethane is burned in excess oxygen, 26.96 g of water
will be produced.
4
Question 6
Question
A gaseous compound containing only carbon and hydrogen is burned completely
in oxygen gas. If 2.00 L of the compound at STP produces 4.40 g of carbon
dioxide and 2.01 g of water, determine the empirical formula of the compound.
Solution
Step 1: Determine the moles of carbon dioxide produced. Given that 4.40 g of
carbon dioxide is produced, we can calculate the moles using the molar mass of
carbon dioxide (CO2). The molar mass of CO2is 44.01 g/mol.
Moles of CO2=4.40 g
44.01 g/mol = 0.1 mol
Step 2: Determine the moles of water produced. Given that 2.01 g of water
is produced, we can calculate the moles using the molar mass of water (H2O).
The molar mass of H2O is 18.02 g/mol.
Moles of H2O = 2.01 g
18.02 g/mol ≈0.1116 mol
Step 3: Calculate the moles of carbon and hydrogen atoms in the compound.
Since the compound contains only carbon and hydrogen, we can assume all of
the carbon in the compound ends up in the CO2and all of the hydrogen ends
up in the H2O. Thus, the moles of carbon and hydrogen can be calculated as
follows:
Moles of C = Moles of CO2= 0.1 mol
Moles of H = 2 ×Moles of H2O=2×0.1116 mol = 0.2232 mol
Step 4: Determine the empirical formula. To find the empirical formula, we
must determine the ratio of moles of carbon to moles of hydrogen. Dividing the
moles of each element by the smallest number of moles (0.1 mol), we get:
C:H
1:2.232
Thus, the empirical formula of the compound is CH2.
Question 7
Question
A 2.00 L container holds 2.00 g of nitrogen gas and 3.00 g of argon gas at a
certain temperature. If the mixture exerts a total pressure of 1.50 atm, what is
the partial pressure of each gas?
5
Solution
Step 1: Calculate the number of moles for each gas using the given masses and
molar masses. - The molar mass of nitrogen (N) is 28.02 g/mol and the molar
mass of argon (Ar) is 39.95 g/mol. - Let nN2be the number of moles of nitrogen
gas and nAr be the number of moles of argon gas.
nN2=2.00 g
28.02 g/mol = 0.0714 mol
nAr =3.00 g
39.95 g/mol = 0.0751 mol
Step 2: Calculate the total moles of gas in the container.
ntotal =nN2+nAr = 0.0714 mol + 0.0751 mol = 0.1465 mol
Step 3: Use the ideal gas law to find the partial pressure of each gas. - The
total pressure (Ptotal) is 1.50 atm. - The volume of the container (V) is 2.00
L. - The ideal gas constant (R) is 0.0821 L
·
atm/mol
·
K. - The temperature is
constant, so we can assume the temperature is in Kelvin.
Ptotal = (PN2+PAr) =⇒PN2+PAr = 1.50 atm
Since pressure is directly proportional to the number of moles of gas, we can
express the partial pressures in terms of the number of moles:
PN2=nN2
ntotal
×Ptotal =0.0714 mol
0.1465 mol ×1.50 atm = 0.732 atm
PAr =nAr
ntotal
×Ptotal =0.0751 mol
0.1465 mol ×1.50 atm = 0.768 atm
Therefore, the partial pressure of nitrogen gas is 0.732 atm and the partial
pressure of argon gas is 0.768 atm.
Question 8
Question
A 2.0 L container is filled with propane gas (C3H8) at a pressure of 3.0 atm and
a temperature of 25
°
C. If the propane reacts completely with excess oxygen gas
according to the following balanced equation:
C3H8+ 5O2→3CO2+ 4H2O
Calculate the volume of carbon dioxide (CO2) produced, assuming the re-
action goes to completion.
(Note: Assume all gases behave ideally.)
6
Solution
Step 1: Convert the temperature to Kelvin.
T(K) = T(C) + 273.15
T= 25C+ 273.15 = 298.15 K
Step 2: Use the ideal gas law to find the number of moles of propane.
P V =nRT
n=P V
RT =(3.0 atm ×2.0 L)
(0.0821 L ·atm/mol ·K×298.15 K)
n=6.0
24.51 = 0.244 mol
Step 3: Determine the limiting reactant. - Calculate the number of moles of
oxygen needed:
moles of O2= 5 ×moles of C3H8= 5 ×0.244 = 1.22 mol
Step 4: Calculate the volume of carbon dioxide produced. - From the bal-
anced chemical equation, we see that 1 mole of C3H8produces 3 moles of CO2.
moles of CO2= 3 ×moles of C3H8= 3 ×0.244 = 0.732 mol
- Use the ideal gas law to find the volume of CO2produced:
V=nRT
P=(0.732 mol ×0.0821 L ·atm/mol ·K×298.15 K)
3.0 atm
V=18.87
3.0= 6.29 L
Therefore, the volume of carbon dioxide (CO2) produced is 6.29 L.
Question 9
Question
A gaseous hydrocarbon fuel is composed of 85.0
Solution
Step 1: Calculate the moles of CO2produced. The molar mass of CO2is 44.01
g/mol. Given that the mass of CO2produced is 10.5 g, we have: Number of
moles of CO2=10.5 g
44.01 g/mol = 0.2386 mol.
Step 2: Calculate the moles of carbon in the hydrocarbon fuel. Assume
we have 100 g of the hydrocarbon fuel. Given that the fuel is composed of
7
85.0Mass of carbon = 100 g ×0.85 = 85 g. The molar mass of carbon is 12.01
g/mol. Number of moles of carbon = 85 g
12.01 g/mol = 7.076 mol.
Step 3: Calculate the moles of hydrogen in the hydrocarbon fuel. Given
that the fuel is composed of 15.0Mass of hydrogen = 100 g ×0.15 = 15 g.
The molar mass of hydrogen is 1.008 g/mol. Number of moles of hydrogen =
15 g
1.008 g/mol = 14.88 mol.
Step 4: Determine the empirical formula of the hydrocarbon fuel. The ratio
of moles of carbon to moles of hydrogen in the fuel is approximately 1:2 (to the
nearest whole number).
Therefore, the empirical formula of the hydrocarbon fuel is C1H2.
Step 5: Determine the molecular formula of the hydrocarbon fuel. The
molecular formula of the hydrocarbon fuel can be found using the molar mass
of the empirical formula and the molar mass of the actual compound.
The molar mass of C1H2is 1 ×12.01 + 2 ×1.008 = 14.03 g/mol.
The molar mass of the hydrocarbon fuel can be calculated as follows: Mo Grp molar mass
Empirical Formula molar mass =
3.45×Molar mass of hydrocarbon
10.5×Molar mass of CO2.
Solving for the molar mass of the hydrocarbon gives 42.09 g/mol.
Finally, the molecular formula of the hydrocarbon fuel is C3H6.
Question 10
Question
Given the reaction:
2 C3H8(g) + 7 O2(g)→6 CO2(g) + 8 H2O(g)
If 12.5 grams of propane (C3H8) reacts with excess oxygen gas, what mass of
water will be produced?
Solution
Step 1: Calculate the molar mass of C3H8. The molar mass of C3H8can be
calculated as:
Molar mass = 3 ×MC+ 8 ×MH
where MC= 12.01 g/mol and MH= 1.008 g/mol. Plugging in the values, we
get:
Molar mass = 3 ×12.01 + 8 ×1.008 = 44.10 g/mol
Step 2: Calculate the number of moles of propane. Use the formula:
Moles = Mass
Molar mass
Substitute the values:
Moles = 12.5
44.10 = 0.283 mol
8
Step 3: Use the mole ratio from the balanced equation to find the number of
moles of water produced. From the balanced equation, 2 moles of C3H8produce
8 moles of H2O. So, 0.283 moles of C3H8will produce:
0.283 mol ×8 mol H2O
2 mol C3H8
= 1.13 mol H2O
Step 4: Calculate the mass of water produced. Use the formula:
Mass = Moles ×Molar mass
Substitute the values:
Mass = 1.13 ×18.015 = 20.33 g
Therefore, 20.33 grams of water will be produced when 12.5 grams of propane
reacts with excess oxygen gas.
Question 11
Question
A sample of a gas with a volume of 4.50 L at a pressure of 1.20 atm and a
temperature of 25
°
C is reacted with excess O2to form SO2gas. If the volume
of SO2produced is 3.00 L at 1.20 atm and 25
°
C, what is the balanced chemical
equation for the reaction?
Solution
Step 1: Write the given information in terms of pressure, volume, and temper-
ature. Given: Initial gas: V1= 4.50 L, P1= 1.20 atm, T1= 25
°
C Final gas
(SO2): V2= 3.00 L, P2= 1.20 atm, T2= 25
°
C
Step 2: Use the ideal gas law to find the number of moles for each gas. For
the initial gas: n1=P1V1
RT1
For the final gas (SO2): n2=P2V2
RT2
Step 3: Since the initial gas reacted with excess O2, the moles of the initial
gas should be equal to the moles of the final gas (SO2). n1=n2
Step 4: Given that SO2is formed in the reaction, the balanced chemical
equation could be:
2A+ 2B→C
Where A and B represent the initial gases, and C represents the formed SO2
gas.
Therefore, the balanced chemical equation for the reaction is:
2A+ 2B→C
9
Question 12
Question
A gaseous hydrocarbon undergoes combustion according to the following bal-
anced chemical equation:
C6H14(g)+9O2(g)→6CO2(g)+7H2O(g)
If 15.0 grams of C6H14 react with excess oxygen, what volume of CO2gas will
be produced at STP (Standard Temperature and Pressure)? (1 mole of gas
occupies 22.4 L at STP)
Solution
Step 1: Calculate the moles of C6H14 in 15.0 grams: Given mass of C6H14: 15.0
grams Molar mass of C6H14: 6(12.01) + 14(1.01) = 86.18 g/mol
Moles of C6H14 =15.0 g
86.18 g/mol ≈0.174 mol
Step 2: Determine the moles of CO2produced using the stoichiometry of
the balanced equation: From the balanced equation, 1 mole of C6H14 produces
6 moles of CO2
Moles of CO2= 0.174 mol ×6 mol CO2
1 mol C6H14
= 1.044 mol
Step 3: Convert moles of CO2to volume at STP using the ideal gas law: At
STP, 1 mole of any gas occupies 22.4 L.
Volume of CO2= 1.044 mol ×22.4 L/mol = 23.462 L
Answer: The volume of CO2produced at STP is 23.462 L.
Question 13
Question
When 5.00 L of ethylene gas (C2H4) is reacted with excess oxygen gas, how many
liters of carbon dioxide gas (CO2) are produced at STP (Standard Temperature
and Pressure)? Assume all gases are ideal.
Solution
Step 1: Write the balanced chemical equation for the reaction between ethylene
and oxygen to produce carbon dioxide and water:
C2H4+ 3O2→2CO2+ 2H2O
10
Step 2: Determine the molar ratio of C2H4to CO2from the balanced equa-
tion: 1 mol of C2H4produces 2 mol of CO2
Step 3: Calculate the number of moles of C2H4using the ideal gas law at
STP:
P V =nRT
n=P V
RT =(1 atm)(5.00 L)
0.0821 L atm/mol K(273 K) = 0.196 mol
Step 4: Using the molar ratio from step 2, calculate the number of moles of
CO2produced:
0.196 mol ×2 mol CO2
1 mol C2H4
= 0.392 mol CO2
Step 5: Convert moles of CO2to volume at STP using the ideal gas law:
P V =nRT
V=nRT
P=(0.392 mol)(0.0821 L atm/mol K)(273 K)
1 atm = 8.17 L
Therefore, when 5.00 L of ethylene gas is reacted with excess oxygen gas,
8.17 L of carbon dioxide gas are produced at STP.
Question 14
Question
A gaseous mixture contains 16.0 g of nitrogen gas (N2) and 32.0 g of oxygen
gas (O2). If the gases react completely to form nitrogen monoxide gas (NO),
what is the mass of the product formed? Assume all gases are at the same
temperature and pressure.
Solution
Step 1: Determine the moles of each reactant using the given masses and molar
masses.
The molar mass of nitrogen gas (N2) is 28.02 g/mol, and the molar mass of
oxygen gas (O2) is 32.00 g/mol.
For nitrogen:
moles of N2=16.0 g
28.02 g/mol = 0.571 mol
For oxygen:
moles of O2=32.0 g
32.00 g/mol = 1.0 mol
Step 2: Determine the limiting reactant by looking at the mole ratio from
the balanced chemical equation.
11
The balanced chemical equation for the reaction is:
N2+ O2→2NO
From the equation, 1 mole of nitrogen gas reacts with 1 mole of oxygen gas
to produce 2 moles of nitrogen monoxide gas.
Since the mole ratio is 1:1, oxygen is the limiting reactant because it will be
completely consumed before all the nitrogen reacts.
Step 3: Calculate the mass of the product formed using the limiting reactant.
Using the mole ratio from the balanced chemical equation, we find that the
1.0 mol of oxygen will produce 2.0 mol of nitrogen monoxide.
The molar mass of nitrogen monoxide (NO) is 30.01 g/mol.
Mass of NO produced = 2.0 mol ×30.01 g/mol = 60.02 g
Therefore, the mass of the product formed when 16.0 g of nitrogen gas and
32.0 g of oxygen gas react completely is 60.02 g of nitrogen monoxide gas.
Question 15
Question
A mixture of 5.00 L of propane gas (C3H8) and 8.00 L of oxygen gas (O2)
are mixed in a container at 27
°
C and 1.00 atm pressure. The two gases react
according to the following balanced equation:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Assuming complete combustion of propane, what is the total volume of the
gaseous mixture at the same temperature and pressure after the reaction is
complete?
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
is:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Step 2: Calculate the moles of propane and oxygen present initially. Since
volume, temperature, and pressure are given, we can use the ideal gas law to
calculate the moles of each gas. The ideal gas law is given by:
P V =nRT
Rearranging to solve for moles, we have:
n=P V
RT
12
For propane:
nC3H8=(1.00 atm)(5.00 L)
0.0821 L ·atm/mol ·K·(300 K) = 0.205 moles
For oxygen:
nO2=(1.00 atm)(8.00 L)
0.0821 L ·atm/mol ·K·(300 K) = 0.328 moles
Step 3: Determine the limiting reactant. From the balanced equation, the
stoichiometric ratio of propane to oxygen is 1:5. Therefore, oxygen is the limiting
reactant since there is less oxygen available for the reaction.
Step 4: Calculate the volume of the gaseous mixture after the reaction is
complete. From the balanced equation, 1 mole of propane reacts with 5 moles of
oxygen to produce 3 moles of carbon dioxide and 4 moles of water. Since oxygen
is the limiting reactant with 0.328 moles, we can calculate the moles of carbon
dioxide and water produced: - Moles of CO2: 0.328 moles ×3
5= 0.197 moles -
Moles of H2O: 0.328 moles ×4
5= 0.262 moles
Now we can use the ideal gas law to find the volume of the gaseous mixture
after the reaction:
V=nRT
P
V=(0.197 moles + 0.262 moles)(0.0821 L ·atm/mol ·K·(300 K))
1.00 atm
V= 12.1 L
Therefore, the total volume of the gaseous mixture after the reaction is
complete is 12.1 L.
Question 16
Question
When 32.0 g of magnesium metal (Mg) react with excess hydrochloric acid
(HCl), how many liters of hydrogen gas (H2) at STP are produced?
Solution
Step 1: Write the balanced chemical equation for the reaction between magne-
sium and hydrochloric acid:
Mg + 2HCl →MgCl2+H2
Step 2: Determine the number of moles of magnesium used: Given mass of
magnesium (Mg): 32.0 g Molar mass of Mg: 24.305 g/mol
Number of moles of Mg =32.0 g
24.305 g/mol ≈1.32 mol
13
Step 3: Use the mole ratio from the balanced chemical equation to find the
number of moles of H2: From the balanced equation, 1 mol of Mg produces 1
mol of H2.
Number of moles of H2= 1.32 mol
Step 4: Calculate the volume of hydrogen gas at STP (Standard Tempera-
ture and Pressure): 1 mol of any gas at STP occupies 22.4 L.
Volume of H2gas at STP = 1.32 mol ×22.4 L/mol = 29.57 L
Step 5: Answer: The reaction of 32.0 g of magnesium with excess hy-
drochloric acid will produce 29.57 L of hydrogen gas at STP.
Question 17
Question
A 2.00 L container holds 0.400 mol of nitrogen gas and 0.800 mol of hydrogen
gas at 500 K. If the gases react to form ammonia gas (NH3) according to the
balanced equation:
N2(g)+3H2(g)→2NH3(g)
What is the limiting reactant? How many moles of ammonia gas can be
produced?
Solution
Step 1: Write the balanced chemical equation
N2(g)+3H2(g)→2NH3(g)
Step 2: Calculate the number of moles of ammonia that can be produced
from nitrogen gas For every mole of nitrogen gas, 2 moles of ammonia are
produced. Therefore, the maximum number of moles of ammonia that can be
produced from nitrogen gas is:
0.400 mol ×2 mol NH3
1 mol N2
= 0.800 mol NH3
Step 3: Calculate the number of moles of ammonia that can be produced
from hydrogen gas For every 3 moles of hydrogen gas, 2 moles of ammonia are
produced. Therefore, the maximum number of moles of ammonia that can be
produced from hydrogen gas is:
0.800 mol ×2 mol NH3
3 mol H2
= 0.533 mol NH3
Step 4: Determine the limiting reactant Since nitrogen gas produces a greater
amount of ammonia (0.800 mol) compared to hydrogen gas (0.533 mol), hydro-
gen gas is the limiting reactant.
14
Step 5: Calculate the number of moles of ammonia gas that can be produced
The limiting reactant is hydrogen gas, which produces 0.533 mol of ammonia.
Thus, 0.533 mol of ammonia gas can be produced.
Question 18
Question
A reaction between acetylene (C2H2) and oxygen produces carbon dioxide and
water. If 10.0 L of acetylene gas at STP is reacted with an excess of oxygen,
what volume of carbon dioxide gas is produced at STP?
Solution
Step 1: Write the balanced chemical equation for the reaction between acetylene
and oxygen:
C2H2+ O2→CO2+ H2O
Step 2: Determine the stoichiometry of the reaction. From the balanced
chemical equation, we see that 1 mole of acetylene produces 1 mole of carbon
dioxide gas.
Step 3: Calculate the number of moles of acetylene that react:
Moles of C2H2=Volume of C2H2
Molar volume at STP =10.0 L
22.4 L/mol = 0.4464 moles
Step 4: Determine the volume of carbon dioxide gas produced at STP using
the Ideal Gas Law:
PV = nRT
VCO2=nCO2RT
P=(0.4464 moles)(0.0821 atm ·L/mol ·K)(273 K)
1 atm
VCO2= 10.4 L
Therefore, 10.4 L of carbon dioxide gas is produced at STP when 10.0 L of
acetylene gas is reacted with excess oxygen.
Question 19
Question
A reaction between solid iron and gaseous sulfur produces solid iron(II) sulfide
according to the following balanced chemical equation:
Fe(s)+S2(g)→FeS(s)
If 5.00 g of iron reacts with an excess of sulfur and produces 8.72 g of iron(II)
sulfide, determine the theoretical and percent yields of the reaction.
15
Solution
Step 1: Calculate the molar mass of iron(II) sulfide. The molar mass of FeS is:
Molar mass of FeS = Mass of Fe+Mass of S = 55.85 g/mol+32.07 g/mol = 87.92 g/mol
Step 2: Determine the number of moles of iron used. Given:
Mass of Fe = 5.00 g
Molar mass of Fe = 55.85 g/mol
Calculate the number of moles of iron:
Moles of Fe = Mass of Fe
Molar mass of Fe =5.00 g
55.85 g/mol = 0.0894 mol
Step 3: Determine the limiting reactant. From the balanced chemical equa-
tion, the stoichiometry of the reaction is 1:1 between Fe and FeS. Therefore, the
limiting reactant is Fe.
Step 4: Calculate the theoretical yield of iron(II) sulfide. Using the mole
ratio from the balanced chemical equation, the theoretical yield can be calcu-
lated:
Moles of FeS produced = Moles of Fe = 0.0894 mol
Calculate the mass of FeS produced:
Mass of FeS produced = Moles of FeS×Molar mass of FeS = 0.0894 mol×87.92 g/mol = 7.86 g
Step 5: Calculate the percent yield of the reaction. Given:
Actual yield = 8.72 g
Theoretical yield = 7.86 g
Calculate the percent yield:
Percent yield = Actual yield
Theoretical yield ×100% = 8.72 g
7.86 g ×100% = 110.89%
Therefore, the theoretical yield of iron(II) sulfide is 7.86 g and the percent
yield of the reaction is 110.89
Question 20
Question
A chemical reaction takes place in a sealed container. The reaction involves the
combustion of propane gas (C3H8) with oxygen gas to produce carbon dioxide
gas and water vapor. If 4.50 L of propane gas reacts with excess oxygen gas at
STP (standard temperature and pressure), what volume of carbon dioxide gas
is produced? Assume all gases are ideal.
16
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the combustion of propane is:
C3H8+ 5O2→3CO2+ 4H2O
Step 2: Determine the moles of propane gas. Using the ideal gas law equation
P V =nRT , we can find the moles of propane gas. Given: Volume (V) = 4.50
L Temperature (T) = 273 K (STP) Pressure (P) = 1 atm Gas constant (R) =
0.0821 L atm/(mol K)
From the ideal gas law:
n=P V
RT =(1 atm)(4.50 L)
(0.0821 L atm/(mol K))(273 K)
Calculating ngives:
n≈0.184 moles
Step 3: Use the stoichiometry of the reaction to find the volume of carbon
dioxide gas produced. From the balanced equation, we know that 1 mole of
propane produces 3 moles of carbon dioxide gas. Therefore, 0.184 moles of
propane produces:
(0.184 moles propane) ×3 moles CO2
1 mole C3H8= 0.552 moles CO2
Now, using the ideal gas law again with the new number of moles, we can
calculate the volume of carbon dioxide gas produced:
V=nRT
P=(0.552 moles)(0.0821 L atm/(mol K))(273 K)
1 atm
Calculating Vgives:
V≈12.8 L
Therefore, approximately 12.8 L of carbon dioxide gas is produced during
the reaction.
Question 21
Question
Calculate the volume of carbon monoxide gas (CO) produced at 273 K and 1
atm when 10.0 g of carbon (C) reacts with excess carbon dioxide gas (CO2)
according to the following balanced chemical equation:
C(s) + CO2(g)→2CO(g)
17
Solution
Step 1: Calculate the number of moles of carbon (C) used.
Molar mass of C= 12.01 g/mol
Moles of C=10.0 g
12.01 g/mol
Moles of C= 0.832 mol
Step 2: Use stoichiometry to find the number of moles of carbon monoxide
gas (CO) produced.
From the balanced equation: 1 mol C→2 mol CO
0.832 mol C→0.832 mol ×2
0.832 mol C→1.664 mol CO
Step 3: Convert moles of carbon monoxide gas to volume at 273 K and 1
atm using the ideal gas law equation P V =nRT .
P= 1 atm
V=?
n= 1.664 mol
R= 0.0821 L ·atm/mol ·K
T= 273 K
Step 4: Solve for the volume of carbon monoxide gas (CO).
V=nRT
P
V=(1.664 mol)(0.0821 L ·atm/mol ·K)(273 K)
1 atm
V= 37.1 L
Therefore, 37.1 L of carbon monoxide gas (CO) is produced.
Question 22
Question
A gaseous compound contains only sulfur and oxygen. When 6.25 g of the
compound is decomposed, 2.35 g of sulfur and 3.90 g of oxygen are produced.
What is the formula of the compound?
(Note: The molar masses are as follows: S = 32.07 g/mol, O = 16.00 g/mol)
18
Solution
Step 1: Calculate the number of moles of sulfur and oxygen produced.
Moles of Sulfur = 2.35 g
32.07 g/mol = 0.0732 mol
Moles of Oxygen = 3.90 g
16.00 g/mol = 0.2438 mol
Step 2: Determine the mole ratio of sulfur to oxygen.
Mole ratio (Sulfur : Oxygen) = 0.0732 mol
0.2438 mol = 0.3:1
Step 3: Find the empirical formula of the compound. Since the formula
contains only sulfur and oxygen, the empirical formula of the compound is SO3.
Therefore, the formula of the compound is sulfur trioxide (SO3).
Question 23
Question
A sample of butane gas (C4H10) is burned in excess oxygen gas (O2) to produce
carbon dioxide gas (CO2) and water vapor (H2O). If 38.0 g of butane reacts
with 280.0 g of oxygen gas, what is the limiting reactant and how many grams
of water vapor are produced?
Solution
Step 1: Write the balanced chemical equation for the reaction:
2C4H10 + 13O2→8CO2+ 10H2O
Step 2: Calculate the molar mass of each substance:
Molar mass of C4H10 = 4(12.01) + 10(1.01) = 58.12 g/mol
Molar mass of O2= 2(16.00) = 32.00 g/mol
Molar mass of CO2= 12.01 + 2(16.00) = 44.01 g/mol
Molar mass of H2O= 2(1.01) + 16.00 = 18.02 g/mol
Step 3: Calculate the number of moles of each reactant:
Moles of C4H10 =38.0 g
58.12 g/mol = 0.654 mol
Moles of O2=280.0 g
32.00 g/mol = 8.75 mol
19
Step 4: Determine the limiting reactant by comparing the mole ratios from
the balanced chemical equation:
Moles of C4H10∇ · 2=0.327 mol
Moles of O2∇ · 13 = 0.673 mol
Since there is more of the butane, it is in excess and oxygen is the limiting
reactant.
Step 5: Calculate the theoretical yield of water vapor:
Moles of water vapor = 8.75 mol ×10
13 = 6.73 mol
Mass of water vapor = 6.73 mol ×18.02 g/mol = 121.0 g
Therefore, the limiting reactant is oxygen gas and 121.0 grams of water vapor
are produced.
Question 24
Question
A sample of a gaseous hydrocarbon undergoes combustion in oxygen gas, pro-
ducing carbon dioxide and water vapor. If 6.40 L of carbon dioxide are produced
at STP, how many moles of the hydrocarbon were in the original sample?
Solution
Step 1: Write the balanced chemical equation for the combustion of the hydro-
carbon. The general formula for a hydrocarbon is CxHy. The balanced chemical
equation for the combustion of a hydrocarbon is:
CxHy+ (x+y/4)O2→xCO2+ (y/2)H2O
Step 2: Calculate the volume of carbon dioxide produced. At STP (Standard
Temperature and Pressure), 1 mole of any gas occupies 22.4 L. The volume of
carbon dioxide produced is 6.40 L. We can use this information to find the
number of moles of carbon dioxide produced:
Moles of CO2=V olume
22.4=6.40
22.4= 0.285 moles
Step 3: Determine the moles of the hydrocarbon. From the balanced chem-
ical equation, we see that the mole ratio of CxHyto CO2is 1:1. Hence, the
moles of the hydrocarbon are also 0.285 moles.
Step 4: Answer Therefore, there were 0.285 moles of the hydrocarbon in the
original sample.
20
Question 25
Question
A 2.50 L container at 27.0
°
C contains 3.00 moles of nitrogen gas and 2.00 moles
of hydrogen gas. The hydrogen gas reacts with the nitrogen gas to form ammonia
gas and no hydrogen is left over. After the reaction, the temperature of the
mixture is 37.0
°
C. What is the pressure in the container after the reaction?
Assume all gases are ideal.
Solution
Step 1: Calculate the initial pressure of the gases using the ideal gas law:
P V =nRT
For nitrogen gas:
PN2=nN2RT
V=(3.00 mol)(0.0821 L ·atm/mol ·K)(27.0 + 273.15 K)
2.50 L
PN2≈6.35 atm
For hydrogen gas:
PH2=nH2RT
V=(2.00 mol)(0.0821 L ·atm/mol ·K)(27.0 + 273.15 K)
2.50 L
PH2≈4.24 atm
Step 2: Write the balanced chemical equation:
3 H2(g)+N2(g)→2 NH3(g)
Step 3: Determine the limiting reactant. Since the hydrogen gas is com-
pletely consumed and nitrogen gas is in excess, nitrogen gas is the limiting
reactant.
Step 4: Calculate the moles of ammonia produced using the stoichiometry of
the reaction: Moles of ammonia produced = 3.00 mol N2
1×2 mol NH3
1 mol N2= 6.00 moles
NH3
Step 5: Calculate the final pressure of the mixture using the ideal gas law:
PNH3=nNH3RT
V=(6.00 mol)(0.0821 L ·atm/mol ·K)(37.0 + 273.15 K)
2.50 L
PNH3≈13.31 atm
Thus, the pressure in the container after the reaction is approximately 13.31
atm.
21
Question 26
Question
A gas mixture contains 6.0 moles of oxygen gas and 4.0 moles of hydrogen gas.
If the gases react according to the equation 2H2(g) + O2(g) →2H2O(g), what
is the limiting reagent?
Solution
Step 1: Write and balance the chemical equation. The balanced chemical equa-
tion is:
2H2(g) + O2(g) →2H2O(g)
This equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen
gas to produce 2 moles of water vapor.
Step 2: Determine the mole ratio of reactants. From the balanced equation,
the mole ratio of H2to O2is 2:1. This means that for every 2 moles of H2, 1
mole of O2is needed to react completely.
Step 3: Calculate the moles of O2needed for the reaction with the available
moles of reactants. For 6.0 moles of oxygen gas, we need (6.0 moles O2)×
(2 moles H2/1 mole O2) = 12.0 moles H2.
Step 4: Compare the moles of O2needed with the moles of H2available.
Since there are only 4.0 moles of hydrogen gas available, which is less than the
12.0 moles required, hydrogen gas is the limiting reagent.
Therefore, hydrogen gas is the limiting reagent in this reaction.
Question 27
Question
A gaseous compound consists of nitrogen and oxygen. If 5.00 L of the compound
at STP decomposes to form 2.00 L of nitrogen gas and the remaining volume is
oxygen gas, determine the empirical formula of the compound. Assume that all
gases are ideal.
Solution
Step 1: Identify the gases involved in the reaction. Let’s denote the compound
as XnOm, where X represents an element and n and m are integers.
Step 2: Determine the moles of nitrogen gas produced. From the ideal gas
law, we can determine the number of moles of nitrogen gas:
molN2=P V
RT =(1.00 atm ×2.00 L)
(0.0821 L ·atm/mol ·K×273 K) = 0.0917 mol
22
Step 3: Determine the moles of oxygen gas produced. Since we know the
total volume is 5.00 L and that 2.00 L is nitrogen gas, the remaining volume
must be oxygen gas:
VolO2= 5.00 L −2.00 L = 3.00 L
Now we can determine the moles of oxygen gas:
molO2=P V
RT =(1.00 atm ×3.00 L)
(0.0821 L ·atm/mol ·K×273 K) = 0.137 mol
Step 4: Determine the ratio of moles of nitrogen and oxygen. Dividing the
moles of oxygen gas by the moles of nitrogen gas gives us the empirical formula:
molO2
molN2
=0.137
0.0917 = 1.49 ≈3
2
Therefore, the empirical formula of the compound is N2O3.
Question 28
Question
A reaction occurs between nitrogen dioxide gas (N O2) and water vapor (H2O) to
produce nitric acid (HNO3) and nitrogen monoxide gas (NO). If 5.00 L of NO2
react with an excess of water vapor at standard temperature and pressure, how
many liters of nitrogen monoxide gas will be produced based on the following
balanced chemical equation?
3NO2(g) + H2O(g)→2HNO3(aq) + N O(g)
Solution
Step 1: Write down the balanced chemical equation for the reaction.
3NO2(g) + H2O(g)→2HNO3(aq) + N O(g)
Step 2: Determine the stoichiometry of the reaction. From the balanced
equation, we see that for every 3 moles of N O2that react, 1 mole of N O is
produced.
Step 3: Convert the volume of NO2to moles using the ideal gas law. Given
that 1 mole of any gas occupies 22.4 L at standard temperature and pressure
(STP), we have:
5.00L×1 mole
22.4L= 0.2232 moles of NO2
Step 4: Determine the volume of NO produced using the mole ratio from
the balanced equation. Since 3 moles of NO2produce 1 mole of NO, we have:
0.2232 moles NO2×1 mole NO
3 moles NO2
= 0.0744 moles NO
23
Step 5: Convert moles of NO to volume at STP. Using the ideal gas law, we
find:
0.0744 moles NO ×22.4L= 1.67 L NO
Answer: The reaction will produce 1.67 liters of nitrogen monoxide gas.
Question 29
Question
A gaseous compound contains only carbon and hydrogen. Combustion of a
0.750 g sample of the compound produced 2.274 g of carbon dioxide and 1.165
g of water. Determine the empirical formula of the compound.
(Assume that all measurements are accurate, and the combustion is com-
plete.)
Solution
Step 1: Calculate the moles of carbon dioxide produced. Given: Mass of carbon
dioxide produced = 2.274 g Molar mass of carbon dioxide (CO2) = 44.01 g/mol
Moles of CO2=Mass of CO2
Molar mass of CO2
Moles of CO2=2.274 g
44.01 g/mol
Moles of CO2≈0.0517 mol
Step 2: Calculate the moles of water produced. Given: Mass of water pro-
duced = 1.165 g Molar mass of water (H2O) = 18.02 g/mol
Moles of H2O = Mass of H2O
Molar mass of H2O
Moles of H2O = 1.165 g
18.02 g/mol
Moles of H2O≈0.0647 mol
Step 3: Determine the moles of carbon and hydrogen in the compound. From
the balanced chemical equation for the combustion reaction of the compound,
we know that 1 mole of C will produce 1 mole of CO2and that 1 mole of H2O
will produce 1 mole of H2O.
Therefore, the ratio of moles of carbon to moles of hydrogen in the compound
is:
Moles of C : Moles of H = 0.0517 mol
0.0647 mol
Simplify the ratio:
Moles of C : Moles of H ≈0.799 : 1
24
Step 4: Determine the empirical formula of the compound. Since the ratio of
moles of carbon to moles of hydrogen is approximately 0.799 : 1, the empirical
formula of the compound is C4H5.
Question 30
Question
A mixture of ethane (C2H6) and oxygen gas (O2) reacts according to the fol-
lowing balanced chemical equation:
C2H6(g) + 7
2O2(g)→2CO2(g) + 3H2O(g)
If 3.00 grams of ethane are reacted with excess oxygen, what mass of water
would be produced?
Solution
Step 1: Calculate the molar mass of ethane (C2H6):
Molar mass of C2H6= 2 ×molar mass of C + 6 ×molar mass of H
= 2 ×12.01 g/mol + 6 ×1.01 g/mol = 30.07 g/mol
Step 2: Calculate the number of moles of ethane used:
Moles of C2H6=Mass
Molar mass =3.00 g
30.07 g/mol = 0.0998 mol
Step 3: Determine the limiting reactant: In this case, ethane is the limiting
reactant because it is given that there is excess oxygen.
Step 4: Calculate the theoretical yield of water produced using the mole
ratio from the balanced chemical equation:
Moles of H2O = Moles of C2H6×3 mol H2O
1 mol C2H6
= 0.0998 mol×3 mol H2O
1 mol C2H6
= 0.299 mol
Step 5: Calculate the mass of water produced:
Mass of H2O = Moles×Molar mass of H2O = 0.299 mol×(2×1.01+16.00) g/mol = 9.02 g
Therefore, 9.02 grams of water would be produced when 3.00 grams of ethane
are reacted with excess oxygen.
25
Question 31
Question
A gaseous compound containing only nitrogen and oxygen is decomposed by
passing an electric current through it. If 37.46 mL of nitrogen gas at 25
°
C and
0.987 atm is produced and collected over water at 25
°
C, what volume of oxygen
gas at 25
°
C and 0.987 atm is collected, assuming the vapor pressure of water
at 25
°
C is 23.8 mmHg?
Solution
Step 1: Calculate the pressure of the gas collected over water. Given that the
vapor pressure of water at 25
°
C is 23.8 mmHg, we need to convert this to atm
to use it in calculations. 1 atm = 760 mmHg, therefore 23.8 mmHg = 23.8/760
atm = 0.03132 atm.
The pressure of the gas collected over water is 0.987 atm (given) −0.03132
atm = 0.95568 atm.
Step 2: Calculate the number of moles of nitrogen gas produced. Using the
ideal gas law P V =nRT , we can calculate the number of moles of nitrogen gas
produced. n1=P V
RT , where Pis the pressure, Vis the volume, Ris the ideal
gas constant, and Tis the temperature in Kelvin. R= 0.0821 L atm/mol K,
T= 25 + 273.15 K.
n1=(0.987)(0.03746/1000)
(0.0821)(298.15)
n10.00000177 mol.
Step 3: Determine the mole ratio between nitrogen and oxygen. The bal-
anced chemical equation for the decomposition of the compound will help us
establish the mole ratio between the gases.
Step 4: Calculate the volume of oxygen gas collected. Since oxygen gas is
twice the number of moles of nitrogen gas, the volume of oxygen gas collected will
be twice that of nitrogen gas. 0.00000177 mol of oxygen gas is produced. Using
the ideal gas law P V =nRT :V2=n2RT
PV2=(2 ×0.00000177) ×(0.0821) ×(298.15)
0.95568
V2=0.00000354 ×24.5585
0.95568
V20.0906 L ≈90.6 mL.
Therefore, approximately 90.6 mL of oxygen gas at 25
°
C and 0.987 atm is
collected.
26
Question 32
Question
A certain amount of propane gas (C3H8) is burned in air. The reaction is as
follows:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
If 25.0 g of propane are burned, what volume of CO2gas is produced at a
temperature of 298 K and a pressure of 1.00 atm?
(Molar mass of C3H8= 44.1 g/mol, CO2= 44.0 g/mol, H2O= 18.0 g/mol,
1 mol CO2at STP occupies 22.4 L)
Solution
Step 1: Convert the mass of propane (C3H8) to moles using its molar mass.
Moles of C3H8=25.0 g
44.1 g/mol
= 0.567 mol
Step 2: Use the stoichiometry of the reaction to find the moles of CO2
that will be produced. Since the balanced equation shows that 1 mole of C3H8
produces 3 moles of CO2, we have:
Moles of CO2= 0.567 mol ×3 mol CO2
1 mol C3H8
= 1.70 mol
Step 3: Calculate the volume of CO2gas produced using the ideal gas law.
The ideal gas law is:
P V =nRT
Given: - n= 1.70 mol, - T= 298 K, - P= 1.00 atm, - R= 0.0821 atm ·
L/mol ·K.
First, convert the temperature to kelvin:
T= 298 K
Now, plug in the values and solve for V:
V=nRT
P
=(1.70 mol)(0.0821 atm ·L/mol ·K)(298 K)
1.00 atm
= 41.7 L
Therefore, the volume of CO2gas produced is 41.7 L.
27
Question 33
Question
When sodium azide (NaN3) is heated, it decomposes into sodium metal and
nitrogen gas. If 10.0 grams of sodium azide is heated, what mass of nitrogen
gas is produced? (Assume all reactants are used up in the reaction.)
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the decomposition of sodium azide is:
2NaN3→2Na + 3N2
Step 2: Calculate the molar mass of NaN3. The molar mass of NaN3:
M(Na) = 22.99 g/mol
M(N) = 14.01 g/mol
M(N3)=3×M(N) = 3 ×14.01 = 42.03 g/mol
M(NaN3) = M(Na) + M(N3) = 22.99 + 42.03 = 65.02 g/mol
Step 3: Calculate the number of moles of NaN3in 10.0 grams.
Number of moles = Mass
Molar mass =10.0 g
65.02 g/mol ≈0.153 mol
Step 4: Use the mole ratio from the balanced equation to calculate the
number of moles of N2produced. From the balanced equation, the mole ratio
of NaN3to N2is 2:3.
Moles of N2= 0.153 mol ×3 mol N2
2 mol NaN3
= 0.229 mol
Step 5: Calculate the mass of N2produced. Using the molar mass of nitrogen
gas (N2):
M(N2)=2×M(N) = 2 ×14.01 = 28.02 g/mol
Mass of N2= Moles ×Molar mass = 0.229 mol ×28.02 g/mol ≈6.42 g
Therefore, when 10.0 grams of sodium azide is heated, approximately 6.42
grams of nitrogen gas is produced.
Question 34
Question
A gaseous compound contains only carbon and hydrogen. When 0.500 g of the
compound is completely burned in oxygen, 1.18 g of carbon dioxide and 0.482
g of water are produced. Determine the empirical formula of the compound.
28
Solution
Step 1: Determine the number of moles of carbon dioxide and water produced.
moles of CO2=1.18 g
44.01 g/mol
= 0.0268 mol
moles of H2O=0.482 g
18.015 g/mol
= 0.0268 mol
Step 2: Determine the moles of carbon and hydrogen in the original com-
pound using the mole ratio in the balanced chemical equation.
moles of C= moles of CO2
= 0.0268 mol
moles of H= 2 ×moles of H2O
= 2 ×0.0268 mol
= 0.0536 mol
Step 3: Determine the empirical formula of the compound.
Molar ratio of C:H=0.0268 mol
0.0536 mol
= 1 : 2
Therefore, the empirical formula of the compound is CH2.
Question 35
Question
If 4.50 L of carbon monoxide gas reacts with 3.25 L of oxygen gas at the same
conditions of temperature and pressure, what volume of carbon dioxide gas
would be produced according to the following balanced chemical equation?
2CO(g)+O2(g)→2CO2(g)
Solution
Step 1: Write down the balanced chemical equation.
CO(g) + 1
2O2(g)→CO2(g)
29
Step 2: Identify the given volumes of gases and their coefficients in the
balanced chemical equation. Given: - Volume of CO: 4.50 L - Volume of O2:
3.25 L From the balanced chemical equation: - 1 mol CO reacts with 1/2 mol
O2to produce 1 mol CO2- Therefore, the volume ratio is 1:1/2:1
Step 3: Convert the volumes of gases to moles using the ideal gas law V=
nRT /P .
CO : nCO =P×VCO
RT =VCO
22.4=4.50 L
22.4 L/mol = 0.200 mol
O2:nO2=P×VO2
RT =VO2
22.4=3.25 L
22.4 L/mol = 0.145 mol
Step 4: Determine the limiting reactant. Since 1 mol of CO reacts with 1/2
mol of O2, and we have 0.200 mol of CO and 0.145 mol of O2, O2is the limiting
reactant.
Step 5: Calculate the volume of CO2gas produced using the limiting reac-
tant. From the balanced chemical equation, 1 mol of CO2is produced per 1/2
mol of O2consumed.
CO2:nCO2= 2 ×(0.145 mol) = 0.290 mol
VCO2=nCO2×22.4=0.290 mol ×22.4 L/mol = 6.50 L
Therefore, the volume of carbon dioxide gas produced would be
6.50 L.
30
