CHEM 131 - ADVANCED GENERAL CHEMISTRY I Gas Stoichiometry Question Bank Set 5

CHEM 131 - ADVANCED GENERAL

CHEMISTRY I - Gas Stoichiometry

Question Bank - Set 5

Liberty University

Question 1

Question

Suppose a gas mixture contains carbon monoxide (CO) and oxygen (O2) in a

3 : 2 molar ratio. If the total pressure of the mixture is 3.00 atm and the

temperature is 25◦C, what is the partial pressure of each gas in the mixture?

Solution

Step 1: Calculate the total number of moles of gas in the mixture.

Total moles = moles of CO + moles of O2

= 3x+ 2x

= 5x

where xrepresents the number of moles of the common factor between CO and

O2.

Step 2: Use the ideal gas law to find the volume occupied by the gas mixture.

P V =nRT

3.00 atm ×V= 5x×0.0821 L atm/mol K ×(25 + 273) K

V=5x×0.0821 ×298

3.00

V= 12.39xL

Step 3: Since the gases are in a 3 : 2 molar ratio, the partial pressures will

also be in a 3 : 2 ratio. Let the partial pressure of CO be 3yand that of O2be

2y.

Step 4: Use the partial pressures to find the moles of each gas.

Moles of CO = 3y×12.39x

0.0821 ×298

Moles of O2=2y×12.39x

0.0821 ×298

Step 5: Set up a system of equations using the molar ratio.

moles of CO

moles of O2

3y×12.39x

0.0821 ×298∇ · 2y×12.39x

0.0821 ×298 =3

3 = 3

Step 6: There seems to be a mistake in the calculations. Let’s reevaluate

the partial pressures and moles of each gas. Reevaluating, we find:

Moles of CO = 3y×12.39x∇ · (0.0821 ×298)

Moles of O2= 2y×12.39x∇ · (0.0821 ×298)

Step 7: Now we can solve the system of equations to find the partial pressures

of each gas.

3y×12.39x

0.0821 ×298 = 3

y=3×0.0821 ×298

3×12.39

y= 6.32 atm

Therefore, the partial pressure of carbon monoxide is 3 ×6.32 = 18.96 atm

and the partial pressure of oxygen is 2 ×6.32 = 12.64 atm.

Question 2

Question

When 10.0 L of butane gas (C4H10) is combusted with excess oxygen gas, how

many liters of carbon dioxide gas will be produced at STP (Standard Temper-

ature and Pressure)? Assume that the combustion of butane is complete.

Solution

Step 1: Write the balanced chemical equation for the combustion of butane:

2C4H10(g) + 13O2(g)→8CO2(g) + 10H2O(g)

Step 2: Calculate the number of moles of butane gas (using ideal gas law

P V =nRT ): Given: Volume of butane gas, V1= 10.0 L Temperature, T=

273 K (STP) Pressure, P= 1 atm Gas constant, R= 0.0821 L ·atm/mol ·K

Using the ideal gas law:

n1=P V

RT =(1 atm)(10.0 L)

(0.0821 L ·atm/mol ·K)(273 K)

n1=10.0

22.37 = 0.447 moles

Step 3: Use the mole ratio from the balanced chemical equation to find

the number of moles of carbon dioxide produced: From the balanced chemical

equation, 2 moles of C4H10 produce 8 moles of CO2. So, for 0.447 moles of

C4H10, the number of moles of CO2produced will be:

Moles of CO2=0.447 moles C4H10 ×8 moles CO2

2 moles C4H10

= 1.79 moles

Step 4: Calculate the volume of carbon dioxide gas at STP: Given that 1

mole of any gas occupies 22.4 L at STP,

V olume = 1.79 moles ×22.4 L/mole = 40.1 L

Therefore, 40.1 liters of carbon dioxide gas will be produced when 10.0 liters

of butane gas is combusted at STP.

Question 3

Question

A reaction occurs between hydrogen gas (H2) and nitrogen gas (N2) according

to the following balanced chemical equation:

N2(g) + 3H2(g)→2NH3(g)

If 5.00 L of nitrogen gas reacts with excess hydrogen gas at STP (standard

temperature and pressure), what volume of ammonia gas is produced?

Solution

Step 1: Calculate the number of moles of nitrogen gas (N2) using the ideal gas

law:

P V =nRT

Since the reaction occurs at STP, the pressure (P) is 1 atm and the tempera-

ture (T) is 273 K. The volume (V) is 5.00 L. The gas constant (R) is 0.0821

L·atm/mol·K.

n=P V

RT =(1.00 atm)(5.00 L)

(0.0821 L ·atm/mol ·K)(273 K)

n=5.00

22.3743 = 0.2234 moles of N2

Step 2: Determine the limiting reactant and the excess reactant.



From the balanced chemical equation, 1 mol of N2reacts with 3 mol of

H2.



Since we have 0.2234 mol of N2, this corresponds to:

0.2234 mol N2×3 mol H2

1 mol N2

= 0.6702 mol H2



If we had an excess of hydrogen gas, the actual number of moles available

for the reaction would be the limiting reactant amount of moles (0.2234

mol). Anything beyond that amount would be in excess.

Step 3: Calculate the volume of ammonia gas (NH3) produced using the

mole ratio from the balanced equation:



From the balanced equation, 1 mol of N2produces 2 mol of NH3.



Hence, 0.2234 mol of N2will produce:

0.2234 mol N2×2 mol NH3

1 mol N2

= 0.4469 mol NH3



Now, we can use the ideal gas law to find the volume of ammonia gas at

STP:

V=nRT

P=(0.4469 mol)(0.0821 L ·atm/mol ·K)(273 K)

1 atm

V=10.8109

1= 10.81 L NH3

Therefore, the volume of ammonia gas produced when 5.00 L of nitrogen gas

reacts with excess hydrogen gas at STP is 10.81 L.

Question 4

Question

A sample of magnesium metal reacts with excess hydrochloric acid to produce

hydrogen gas and magnesium chloride. If 3.00 grams of magnesium completely

reacts, how many liters of hydrogen gas are produced at STP? (Assume the

molar volume of a gas at STP is 22.4 L/mol)

Solution

Step 1: Write the balanced chemical equation for the reaction between magne-

sium and hydrochloric acid. The balanced chemical equation is:

Mg + 2HCl →MgCl2+ H2

Step 2: Calculate the number of moles of magnesium reacting. Given: mass

of magnesium = 3.00 grams Molar mass of magnesium = 24.31 g/mol

Number of moles of magnesium = 3.00 g

24.31 g/mol = 0.1236 mol

Step 3: Use the stoichiometry of the balanced chemical equation to determine

the moles of hydrogen gas produced. From the balanced equation, 1 mole of

magnesium produces 1 mole of hydrogen gas. Number of moles of hydrogen gas

= 0.1236 mol

Step 4: Calculate the volume of hydrogen gas at STP. Given: Molar volume

of gas at STP = 22.4 L/mol

Volume of hydrogen gas = 0.1236 mol ×22.4 L/mol = 2.77 L

Therefore, 2.77 liters of hydrogen gas are produced at STP when 3.00 grams

of magnesium completely reacts.

Question 5

Question

A gaseous compound containing only carbon and hydrogen is burned in oxygen

gas. The products of the combustion reaction are carbon dioxide and water

vapor. If 1.00 g of the compound produces 2.88 g of carbon dioxide and 1.12 g

of water vapor, what is the empirical formula of the compound?

Solution

Step 1: Calculate the moles of carbon dioxide produced. - The molar mass of

carbon dioxide (CO2) is approximately 44.01 g/mol. - Using the given mass of

carbon dioxide:

moles of CO2=2.88 g

44.01 g/mol = 0.06544 mol

Step 2: Calculate the moles of water vapor produced. - The molar mass of

water (H2O) is approximately 18.02 g/mol. - Using the given mass of water:

moles of H2O = 1.12 g

18.02 g/mol = 0.06215 mol

Step 3: Determine the moles of carbon and hydrogen in the compound. -

In the combustion reaction, carbon from the compound forms carbon dioxide,

while hydrogen forms water vapor. - Assume the compound contains xmoles

of carbon and ymoles of hydrogen. - From the moles of products produced:

x= 0.06544 mol of C, y = 0.12430 mol of H

Step 4: Calculate the ratio of moles of carbon and hydrogen. - To find the

simplest whole number ratio of carbon to hydrogen, we need to divide by the

smallest value of xor y, which is 0.06544 mol.

0.06544 = 1,y

0.06544 ≈1.90

Step 5: Find the empirical formula. - Since the ratio is nearly 2, we multiply

both xand yby 2:

2C: 2H⇒C2H2

Step 6: The empirical formula of the compound is C2H2.

Question 6

Question

A gaseous compound containing only carbon and hydrogen was decomposed

completely through a series of reactions. The volume of carbon dioxide gas

produced was found to be 11.2 L at 298 K and 1 atm. If the gas was originally

2.40 g and had a density of 1.964 g/L at the same temperature and pressure,

determine the molecular formula of the compound.

(Molar volume of a gas at STP = 22.4 L/mol)

Solution

Step 1: Find the molar mass of the compound



Given mass of gas, m= 2.40 g



Molar volume of a gas at STP, Vmolar = 22.4 L/mol



Density of the gas, ρ= 1.964 g/L

Step 2: Calculate the molar mass of the compound

Molar mass = mass

number of moles =ρ×Vmolar

Volume of gas

Molar mass = 1.964 g/L ×22.4 L/mol

2.40 g = 18.27 g/mol

Step 3: Determine the empirical formula of the compound



The empirical formula is the simplest ratio of the elements in the com-

pound

Step 4: Determine the molecular formula of the compound



Given molar mass of compound, Mcompound = 18.27 g/mol

Molecular formula = n×Empirical formula

n=Molecular formula molar mass

Empirical formula molar mass =18.27

Empirical formula molar mass

Since the molecular formula of the compound is twice the empirical formula:

n= 2

Therefore, the molecular formula of the compound is twice the empirical

formula.

Question 7

Question

A sample of zinc metal reacts with hydrochloric acid to produce zinc chloride

and hydrogen gas. If 20.0 grams of zinc reacts with an excess of hydrochloric

acid, how many liters of hydrogen gas are produced at STP?

Given: Molar volume of gas at STP = 22.4 L/mol

Solution

Step 1: Write the balanced chemical equation for the reaction.

Zn + 2HCl →ZnCl2+ H2

Step 2: Calculate the number of moles of zinc involved in the reaction.

Molar mass of Zn = 65.38 g/mol

Moles of Zn = 20.0 g

65.38 g/mol

= 0.306 mol

Step 3: Determine the number of moles of hydrogen gas produced using the

mole ratio from the balanced chemical equation.

Moles of H2= 0.306 mol ×1 mol H2

1 mol Zn

= 0.306 mol

Step 4: Calculate the volume of hydrogen gas produced at STP.

Volume of H2= 0.306 mol ×22.4 L/mol

= 6.86 L

Answer: The volume of hydrogen gas produced at STP is 6.86 liters.

Question 8

Question

A gas mixture contains 2.0 moles of oxygen gas (O2) and 3.0 moles of nitrogen

gas (N2). If the temperature and pressure are held constant, what is the total

volume of the gas mixture at standard temperature and pressure (STP)? Assume

ideal gas behavior.

Solution

Step 1: Write the balanced chemical equation for the reaction of oxygen and

nitrogen gases. Step 2: Determine the mole ratio of oxygen to nitrogen in the

reaction. Step 3: Use the ideal gas law to calculate the total volume of the gas

mixture at STP.

Step 1: Write the balanced chemical equation for the reaction of oxygen and

nitrogen gases. The balanced chemical equation for the reaction of oxygen and

nitrogen gases is:

2O2(g) + N2(g)→2O2N2(g)

Step 2: Determine the mole ratio of oxygen to nitrogen in the reaction. From

the balanced chemical equation, the mole ratio of oxygen to nitrogen is 2:1.

Step 3: Use the ideal gas law to calculate the total volume of the gas mixture

at STP. At STP (standard temperature and pressure), the conditions are: -

Temperature = 273 K - Pressure = 1 atm - Volume of 1 mole of gas at STP =

22.4 L

Since the mole ratio of oxygen to nitrogen is 2:1, we can find the moles of

oxygen and nitrogen in the mixture: - Moles of oxygen: 2.0 moles - Moles of

nitrogen: 3.0 moles

Now, we can calculate the total volume of the gas mixture using the ideal

gas law:

P V =nRT

V=nRT

V=(2.0+3.0) ×0.0821 ×273

V=5.0×0.0821 ×273

V=112.27

V≈112.27 L

Therefore, the total volume of the gas mixture at STP is approximately

112.27 L.

Question 9

Question

In a reaction involving the combustion of butane gas (C4H10), 12 moles of

butane gas react with excess oxygen gas to produce carbon dioxide and water

vapor. Calculate the volume of carbon dioxide produced at STP.

Solution

Step 1: Write the balanced chemical equation for the combustion of butane:

2C4H10 + 13O2→8CO2+ 10H2O

Step 2: Determine the molar ratio between butane and carbon dioxide from

the balanced equation: 1 mole of C4H10 reacts to produce 8 moles of CO2

Step 3: Calculate the number of moles of carbon dioxide produced when 12

moles of butane react:

12 moles of C4H10 ×8 moles of CO2

2 moles of C4H10

= 48 moles of CO2

Step 4: Convert moles of carbon dioxide to volume at STP (standard tem-

perature and pressure, which is 0

C and 1 atm pressure):

48 moles of CO2×22.4 L/mol = 1075.2 L of CO2

Therefore, 1075.2 L of carbon dioxide is produced during the combustion of

12 moles of butane at STP.

Question 10

Question

A 4.00 L flask contains gas at a pressure of 3.00 atm and a temperature of 25

This gas is then transferred to a 2.00 L flask at 25

C. What is the new pressure

inside the 2.00 L flask?

(Hint: Remember to use the ideal gas law, P V =nRT , where Pis pressure,

Vis volume, nis the number of moles of gas, Ris the gas constant, and Tis

temperature in Kelvin.)

Solution

Step 1: Convert the temperature from Celsius to Kelvin. Given that the

temperature is 25

C, we can convert it to Kelvin using the formula T(K) =

T(C) + 273.15. Therefore, T(K) = 25 + 273.15 = 298.15 K.

Step 2: Calculate the number of moles of gas using the ideal gas law, P V =

nRT . For the initial condition in the 4.00 L flask: P1= 3.00 atm, V1= 4.00 L,

T1= 298.15 K.

n=P1V1

RT1=(3.00 atm)(4.00 L)

0.0821 L atm/mol K·298.15 K n≈0.486 mol.

Step 3: Apply the ideal gas law to find the new pressure in the 2.00 L flask.

Since moles of gas remain constant, we have: P1V1=P2V2.

P2=P1V1

V2=(3.00 atm)(4.00 L)

2.00 L .P2= 6.00 atm.

Therefore, the new pressure inside the 2.00 L flask is 6.00 atm.

Question 11

Question

A gaseous compound containing only carbon and hydrogen was burned in excess

oxygen. The products were carbon dioxide and water vapor. If 0.550 grams of

the compound produced 1.560 grams of carbon dioxide and 0.640 grams of water

vapor, determine the empirical formula of the compound.

Solution

Step 1: Find the moles of each product produced. Given: - Mass of compound

burned = 0.550 g - Mass of carbon dioxide produced = 1.560 g - Mass of water

vapor produced = 0.640 g

Calculate the moles of carbon dioxide produced:

Moles of CO2=1.560 g

44.01 g/mol = 0.0355 mol

Calculate the moles of water vapor produced:

Moles of H2O=0.640 g

18.02 g/mol = 0.0355 mol

Step 2: Find the moles of carbon and hydrogen in the compound. From the

balanced chemical equation for the combustion of the compound:

CaHb+ O2→CO2+ H2O

we see that 1 mole of carbon compound produces 1 mole of CO2and 1 mole of

H2O.

Since the moles of carbon dioxide produced is equal to the moles of water

vapor produced, the moles of carbon produced is also equal to the moles of

hydrogen produced.

Step 3: Find the empirical formula of the compound. Calculate the molar

mass of each product: - CO2: 12.01 + 2(16.00) = 44.01 g/mol - H2O: 2(1.01) +

16.00 = 18.02 g/mol

The molar mass of CO2consists of 1 mole of carbon and 2 moles of oxygen,

while the molar mass of H2Oconsists of 2 moles of hydrogen and 1 mole of

oxygen.

Since carbon and hydrogen are the only elements in the compound, the molar

mass of the compound is approximately:

(12.01a+ 1.01b) g/mol

From the moles of carbon dioxide and water vapor produced, we know that:

0.0355 = a

12.01 =b

1.01

Solving these equations simultaneously, we find: a= 1 and b= 1

Therefore, the empirical formula of the compound is CH.

Question 12

Question

In the combustion of propane (C3H8) with excess oxygen, propane is converted

to carbon dioxide and water vapor according to the following balanced equation:

C3H8+ 5O2→3CO2+ 4H2O

If 20.0 g of propane is burned, how many grams of water vapor are produced?

Solution

Step 1: Calculate the number of moles of propane burned. Given: Mass of

propane, C3H8: 20.0 g Molar mass of C3H8: 3(12.01 g/mol) + 8(1.008 g/mol)

= 44.11 g/mol

Number of moles of C3H8=20.0 g

44.11 g/mol = 0.453 mol

Step 2: Determine the limiting reactant. Using the balanced equation, we

see that 1 mole of C3H8produces 4 moles of H2O. Therefore, 0.453 moles of

C3H8will produce: 0.453 mol C3H8×4 mol H2O

1 mol C3H8

= 1.81 mol H2O

Step 3: Calculate the mass of water vapor produced. Molar mass of H2O:

2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol

Mass of H2Oproduced = 1.81 mol ×18.02 g/mol = 32.6 g

Therefore, 32.6 grams of water vapor are produced when 20.0 grams of

propane is burned.

Question 13

Question

A gaseous compound containing carbon, hydrogen, and oxygen was burned in

excess oxygen gas. The combustion of 0.300 g of this compound produced 0.882

g of carbon dioxide and 0.506 g of water. Determine the empirical formula of

the compound.

Solution

Step 1: Calculate moles of carbon dioxide produced. The molar mass of carbon

dioxide (CO2) is 44.01 g/mol. Therefore,

moles of CO2=0.882 g

44.01 g/mol = 0.02 mol

Step 2: Calculate moles of water produced. The molar mass of water (H2O)

is 18.015 g/mol. Therefore,

moles of H2O = 0.506 g

18.015 g/mol = 0.028 mol

Step 3: Determine the moles of carbon, hydrogen, and oxygen in the com-

pound. Let the moles of carbon, hydrogen, and oxygen in the compound be

represented by x, y, and z respectively. From the chemical equation of the

combustion reaction, we have the following relationships:

1 mole of C produces 1 mole of CO2.

1 mole of H2produces 1/2 moles of H2O.

Using the given information, we can set up the following equations:

x= 0.02 mol of C

y= 2 ×0.028 mol of H

2x+y=z

Step 4: Calculate the molar ratios. The simplest way to proceed is to choose

a common multiplier that will make all the moles whole numbers. By inspection,

we can see that multiplying all values by 50 will give whole numbers:

x= 1 mol of C

y= 2 mol of H

z= 100 mol of O

Step 5: Determine the empirical formula. The empirical formula of the

compound is C1H2O100. Simplifying, we find that the empirical formula is

CHO50.

Question 14

Question

A gaseous compound containing only carbon and hydrogen is burned in excess

oxygen. It is found that 1.00 L of the gas at 0.980 atm and 27.0

C gives 2.17 g

CO2at the same temperature and pressure. What is the molecular formula of

the compound?

(Molar masses: C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol)

Solution

Step 1: Calculate the moles of CO2produced. Given mass of CO2= 2.17 g

Molar mass of CO2= 44.01 g/mol (from the atomic masses of C and O) Number

of moles of CO2=2.17 g

44.01 g/mol = 0.0493 mol

Step 2: Calculate the moles of O2consumed. From the balanced chemi-

cal equation for the combustion of the gaseous compound: 1 mole of gaseous

compound + 3

2moles of O2→2 moles of CO2+ x moles of H2O

Since the gaseous compound contains only carbon and hydrogen, the only

source of carbon in the products is from CO2. Therefore: 1 mole of gaseous

compound →2 moles of CO2

So, moles of O2consumed = 2 ×moles of CO2= 2 ×0.0493 = 0.0986 mol

Step 3: Calculate the moles of gaseous compound. Since the gaseous com-

pound is burned in excess oxygen, all carbon and hydrogen in the compound

end up in the products. Therefore, the moles of gaseous compound consumed in

the reaction is equal to the moles of O2consumed: Moles of gaseous compound

= 0.0986 mol

Step 4: Calculate the molar mass of the gaseous compound. Given moles of

gaseous compound = 0.0986 mol Given volume of gaseous compound = 1.00 L

Given pressure of gaseous compound = 0.980 atm Given temperature of gaseous

compound = 27.0

C = 300.0 K

Using the ideal gas law, P V =nRT , we can rearrange to find molar mass:

M=mRT

P V

Substitute the given values: M=(0.0986 mol)(0.0821 L·atm/mol/K)(300.0 K)

(0.980 atm)(1.00 L)

M=2.4480

0.980 = 2.5 g/mol

Step 5: Determine the empirical formula of the compound. Given the masses

of carbon and hydrogen in the compound, we can find the mole ratio of C to

H in the compound: Molar mass of C = 12.01 g/mol Molar mass of H = 1.01

g/mol

Assume 1 mole of the compound contains x moles of carbon and y moles of

hydrogen: 12.01x + 1.01y = 2.5

Solving for the smallest whole number ratio of x to y gives the empirical

formula of the compound.

Step 6: Calculate the molecular formula of the compound. Use the molar

mass of the compound to determine the molecular formula. For example, if the

empirical formula is CH, and the molar mass is 50 g/mol, then the molecular

formula would be C2H2.

Question 15

Question

A gaseous hydrocarbon reacts with oxygen according to the following balanced

chemical equation:

CnHm+ (n+m

4)O2→nCO2+m

2H2O

If 5.00 g of the hydrocarbon reacts with 50.0 g of oxygen and produces 13.2 g of

carbon dioxide, determine the empirical formula of the hydrocarbon. (Assume

all substances are gases at the conditions of the reaction.)

Solution

Step 1: Convert the masses given to moles. The molar mass of oxygen (O2) is

32.00 g/mol. The molar mass of carbon dioxide (CO2) is 44.01 g/mol. We can

use these molar masses to convert grams to moles. For the hydrocarbon, we do

not know the molar mass, so we’ll denote it as M.

Given: Mass of hydrocarbon = 5.00 g Mass of oxygen = 50.0 g Mass of CO2

produced = 13.2 g

Number of moles of hydrocarbon:

moles of hydrocarbon = 5.00 g

Number of moles of oxygen:

moles of oxygen = 50.0 g

32.00 g/mol

Number of moles of CO2 produced:

moles of CO2 = 13.2 g

44.01 g/mol

Step 2: Use stoichiometry to relate the moles of reactants and products to

find the empirical formula. From the balanced chemical equation, 1 mole of

hydrocarbon reacts with n+m/4 moles of O2to produce nmoles of CO2. This

can be used to set up the following ratio:

moles of hydrocarbon

moles of oxygen =

5.00

50.0

32.00

n+m/4

Similarly, the ratio of moles of hydrocarbon to moles of CO2is:

moles of hydrocarbon

moles of CO2 =

5.00

13.2

44.01

Solving these two equations simultaneously will allow us to determine the

values of nand m, and hence, the empirical formula of the hydrocarbon.

Question 16

Question

A gaseous mixture contains 2.0 moles of oxygen gas (O2) and 3.0 moles of

hydrogen gas (H2). If the mixture reacts completely to form water vapor (H2O),

what is the maximum amount of water vapor that can be produced? Assume

all gases are at the same temperature and pressure.

Solution

Step 1: Write and balance the chemical equation for the reaction. In this case,

the balanced chemical equation is:

2H2+O2→2H2O

Step 2: Determine the limiting reactant. To find the limiting reactant, we

need to calculate the moles of product that can be formed from each reactant.

Then, we compare these amounts to determine which reactant limits the amount

of product that can be produced. From the balanced equation, we see that 2

moles of water (H2O) are produced for every 1 mole of oxygen gas (O2) used,

and 2 moles of water are produced for every 2 moles of hydrogen gas (H2) used.

For oxygen gas (O2): - Moles of water produced = 2 moles of oxygen gas ×

(2 moles of water / 1 mole of oxygen gas) = 4 moles of water

For hydrogen gas (H2): - Moles of water produced = 3 moles of hydrogen

gas ×(2 moles of water / 2 moles of hydrogen gas) = 3 moles of water

Since the amount of water produced from 3 moles of hydrogen gas is less

than the amount of water produced from 2 moles of oxygen gas, hydrogen gas

is the limiting reactant.

Step 3: Calculate the maximum amount of water vapor that can be pro-

duced. Since hydrogen gas is the limiting reactant, we will use the moles of

hydrogen gas to determine the moles of water produced. Using the stoichiome-

try from the balanced equation:

3 moles of H2×(2 moles of H2O/2 moles of H2) = 3 moles of H2O

Therefore, the maximum amount of water vapor that can be produced is 3.0

moles.

Question 17

Question

When 5.00 L of methane gas (CH4) is burned in excess oxygen gas (O2) at STP,

how many liters of carbon dioxide gas (CO2) are produced?

Solution

Step 1: Write the balanced chemical equation for the combustion of methane:

CH4(g) + 2O2(g)→CO2(g) + 2H2O(l)

Step 2: Calculate the number of moles of methane used:

n(CH4) = V(CH4)

Vmolar

=5.00 L

22.4 L/mol = 0.223 mol

Step 3: Determine the number of moles of carbon dioxide produced using

the mole ratio from the balanced chemical equation:

n(CO2) = n(CH4)=0.223 mol

Step 4: Convert the number of moles of carbon dioxide to volume at STP:

V(CO2) = n(CO2)×Vmolar = 0.223 mol ×22.4 L/mol = 5.00 L

Therefore, 5.00 liters of carbon dioxide gas are produced when 5.00 liters of

methane gas is burned in excess oxygen gas at STP.

Question 18

Question

A mixture of 1.00 g hydrogen gas and 8.00 g oxygen gas is ignited to form water.

Calculate the maximum mass of water that can be produced.

Solution

Step 1: Write the balanced chemical equation for the reaction.

2H2+ O2→2H2O

Step 2: Calculate the number of moles of hydrogen and oxygen gas present.

Given: Mass of hydrogen gas (H2) = 1.00 g Molar mass of hydrogen = 1.008

g/mol Mass of oxygen gas (O2) = 8.00 g Molar mass of oxygen = 16.00 g/mol

Number of moles of hydrogen:

moles of H2=1.00 g

1.008 g/mol = 0.9921 mol

Number of moles of oxygen:

moles of O2=8.00 g

16.00 g/mol = 0.5000 mol

Step 3: Determine the limiting reactant. From the balanced chemical equa-

tion, 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.

Therefore, the stoichiometry ratio is 2:1.

Since there is more hydrogen (0.9921 mol) than oxygen (0.5000 mol), oxygen

is the limiting reactant.

Step 4: Calculate the maximum mass of water that can be produced. Using

the molar mass of water (18.02 g/mol):

Max mass of water = 0.5000 mol ×18.02 g/mol = 9.01 g

Therefore, the maximum mass of water that can be produced is 9.01 grams.

Question 19

Question

A reaction takes place according to the following balanced equation:

2 C3H8(g) + 7 O2(g)→6 CO2(g) + 8 H2O(g)

If 4.50 moles of propane (C3H8) and 12.0 moles of oxygen gas are present

initially, determine: (a) the limiting reactant, (b) the theoretical yield of water

in moles.

Solution

(a) To determine the limiting reactant, we need to calculate the moles of water

formed by each reactant and identify which reactant produces the least amount

of water. Let’s start by calculating the moles of water produced by each reactant:

Step 1: Calculate moles of water produced by propane

Given: Moles of propane = 4.50 mol From the balanced equation: Moles of

water produced per mole of propane = 8 mol

Therefore, moles of water produced by propane = 4.50 mol ×8 mol H2O

2 mol C3H8=

18.0 mol H2O

Step 2: Calculate moles of water produced by oxygen

Given: Moles of oxygen = 12.0 mol From the balanced equation: Moles of

water produced per mole of oxygen = 8 mol

Therefore, moles of water produced by oxygen = 12.0 mol ×8 mol H2O

7 mol O2

≈

13.71 mol H2O

Since oxygen produces fewer moles of water compared to propane, oxygen is

the limiting reactant.

(b) To determine the theoretical yield of water in moles, we use the moles

of water produced by the limiting reactant, which is oxygen in this case. Theo-

retical yield of water = 13.71 mol.

Question 20

Question

A gaseous compound containing only carbon and hydrogen is burned in ex-

cess oxygen, producing carbon dioxide and water vapor. If the molar mass of

the compound is 78.11 g/mol and the combustion of 0.750 g of the compound

produces 2.168 g of water, what is the molecular formula of the compound?

Solution

Step 1: Calculate the moles of water produced. Given that 0.750 g of the com-

pound produces 2.168 g of water, we can calculate the moles of water produced

using the molar mass of water (18.015 g/mol).

Moles of water = 2.168 g

18.015 g/mol ≈0.120 mol

Step 2: Calculate the moles of carbon in the compound. From the balanced

combustion reaction of the compound, we know that 1 mol of the compound

produces 1 mol of carbon dioxide. As a result, the moles of carbon produced

are equal to the moles of water produced.

Moles of carbon = 0.120 mol

Step 3: Calculate the moles of hydrogen in the compound. In the combustion

reaction, the moles of hydrogen are found by subtracting the moles of carbon

from the moles of the compound.

Moles of hydrogen = Moles of compound −Moles of carbon

Since the compound only contains carbon and hydrogen, the moles of the com-

pound are the sum of the moles of carbon and hydrogen.

Moles of hydrogen = Moles of compound−Moles of carbon = Moles of carbon = 0.120 mol

Step 4: Calculate the molar ratios of carbon and hydrogen. The molar ratio

of carbon to hydrogen in the compound can be written as:

Molar ratio of carbon to hydrogen = Moles of carbon

Moles of hydrogen =0.120 mol

0.120 mol = 1 : 1

Step 5: Determine the empirical formula of the compound. Since the molar

ratio of carbon to hydrogen is 1:1, the empirical formula of the compound is

CH.

Step 6: Calculate the molecular formula of the compound. To find the

molecular formula, we need to compare the molar mass of the empirical formula

with the given molar mass of the compound. The molar mass of CH is:

MCH = 12.011 g/mol + 1.008 g/mol = 13.019 g/mol

Step 7: Determine the molecular formula. To find the molecular formula, we

calculate the ratio of the given molar mass to the molar mass of the empirical

formula:

Molecular formula ratio = 78.11 g/mol

13.019 g/mol ≈6

Therefore, the molecular formula of the compound is 6 times the empirical

formula:

Molecular formula = 6 ×empirical formula = 6 ×CH = C6H6

Therefore, the molecular formula of the compound is C6H6.

Question 21

Question

A gas mixture contains methane (CH4) and oxygen (O2) in a 3:7 molar ratio. If

2.5 moles of methane react with excess oxygen to produce carbon dioxide and

water, how many moles of oxygen are needed?

Solution

Step 1: Write the balanced chemical equation for the combustion of methane:

CH4+ 2O2→CO2+ 2H2O

Step 2: Determine the moles of oxygen required to react with 2.5 moles of

methane. Since the ratio of CH4to O2is 3:7, we need to use the stoichiometry

to find the moles of O2.

2.5 moles CH4

1×7 moles O2

3 moles CH4

=17.5

3= 5.83 moles O2

Answer: 5.83 moles of oxygen are needed.

Question 22

Question

A gaseous compound contains only carbon and hydrogen. When 2.00 g of this

compound is burned in excess oxygen, 4.44 g of carbon dioxide and 1.80 g of

water are produced. Determine the empirical formula of the compound.

Solution

Step 1: Calculate the moles of carbon dioxide produced. Given: Mass of carbon

dioxide produced = 4.44 g

The molar mass of carbon dioxide (CO2) is approximately 44.01 g/mol.

Therefore, the moles of CO2produced can be calculated as:

Moles of CO2=mass

molar mass =4.44 g

44.01 g/mol

Step 2: Calculate the moles of water produced. Given: Mass of water pro-

duced = 1.80 g

The molar mass of water (H2O) is approximately 18.02 g/mol.

Therefore, the moles of H2O produced can be calculated as:

Moles of H2O = mass

molar mass =1.80 g

18.02 g/mol

Step 3: Determine the moles of carbon and hydrogen in the compound.

Let the moles of carbon be represented as x, and the moles of hydrogen be

represented as y.

From the balanced chemical equation of the combustion reaction, we know

that 1 mole of C produces 1 mole of CO2and 1 mole of H2produces 1 mole of

H2O.

Thus, we have the following relationships:

(x= Moles of CO2

y= Moles of H2O

Step 4: Calculate the mole ratio of carbon to hydrogen. We now have:

(x=4.44

44.01

y=1.80

18.02

Step 5: Simplify the mole ratio to get the empirical formula. Solving for x

and ygives:

(x≈0.101

y≈0.100

Since these are nearly equal, the empirical formula of the compound is CH

or C1H1.

Question 23

Question

A gas mixture contains 2.00 g of methane (CH4), 3.00 g of nitrogen gas (N2),

and 4.00 g of oxygen gas (O2). If the total pressure of the mixture is 3.00 atm,

what is the partial pressure of nitrogen gas in the mixture?

Given: Molar mass of CH4: 16.05 g/mol Molar mass of N2: 28.02 g/mol

Molar mass of O2: 32.00 g/mol

Solution

Step 1: Calculate the number of moles of each gas present in the mixture.

Moles of CH4=2.00 g

16.05 g/mol = 0.124 mol

Moles of N2=3.00 g

28.02 g/mol = 0.107 mol

Moles of O2=4.00 g

32.00 g/mol = 0.125 mol

Step 2: Calculate the mole fraction of nitrogen gas. The mole fraction of a

gas component is given by:

Mole fraction = moles of the component

total moles in the mixture

Calculate the total moles in the mixture:

Total moles = 0.124 mol + 0.107 mol + 0.125 mol = 0.356 mol

Now, calculate the mole fraction of N2:

Mole fraction of N2=0.107 mol

0.356 mol = 0.3006

Step 3: Calculate the partial pressure of nitrogen gas. Since the total pres-

sure of the mixture is 3.00 atm, multiply the mole fraction of nitrogen by the

total pressure to get the partial pressure of nitrogen gas:

Partial pressure of N2= 0.3006 ×3.00 atm = 0.902 atm

Therefore, the partial pressure of nitrogen gas in the mixture is 0.902 atm.

Question 24

Question

A mixture of hydrogen gas and oxygen gas is ignited to produce water vapor.

If 2.00 L of hydrogen gas reacts with excess oxygen gas to produce 3.00 L of

water vapor, what volume of oxygen gas (in L) was present initially? Assume

all gases are at the same temperature and pressure.

Solution

Step 1: Write the balanced chemical equation for the reaction.

2H2(g)+O2(g)→2H2O(g)

Step 2: Determine the mole ratio between hydrogen gas and water vapor.

In the balanced chemical equation, 2 moles of hydrogen gas react to produce

2 moles of water vapor. This gives a 1:1 mole ratio between hydrogen gas and

water vapor.

Step 3: Calculate the moles of hydrogen gas reacting. Given that 2.00 L of

hydrogen gas is used and using the ideal gas law P V =nRT , we can calculate

the moles of hydrogen gas. We know that nH2=P V

RT . Substitute P= constant,

V= 2.00 L, R= 0.0821 L ·atm/mol ·K, and T= constant (since all gases are

at the same temperature and pressure).

Step 4: Calculate the moles of oxygen gas reacting. From the balanced

chemical equation, the mole ratio between hydrogen gas and oxygen gas is 1:1

(since the hydrogen gas reacts completely with excess oxygen). So, the moles

of oxygen gas reacting are equal to the moles of hydrogen gas reacting.

Step 5: Calculate the volume of oxygen gas initially present. Assuming all

the oxygen gas gets used up in the reaction, we can determine the volume of

oxygen gas by using the ideal gas law: P V =nRT . Substitute P= constant,

n= moles of oxygen gas calculated in Step 4, R= 0.0821 L ·atm/mol ·K, and

T= constant.

Question 25

Question

A gaseous compound with the formula XCl decomposes according to the fol-

lowing reaction:

2XCl(g)→2X(g) + Cl2(g)

If 3.50 moles of XCl decomposes completely at a constant pressure and tem-

perature, what volume will the resulting gases occupy? Assume all gases are

ideal.

Solution

Step 1: Write the balanced chemical equation for the reaction.

2XCl(g)→2X(g) + Cl2(g)

Step 2: Determine the stoichiometry of the reaction. According to the bal-

anced equation, for every 2 moles of XCl that decompose, 2 moles of Xand 1

mole of Cl2areproduced.

Step 3: Calculate the number of moles of products produced from the given

amount of reactant. Given: Moles of XCl = 3.50 mol Using the stoichiom-

etry from step 2, we find that 3.50 moles of XCl will produce: - 3.50 mol ×

2 mol X

2 mol XCl = 3.50 mol X - 3.50 mol ×1 mol Cl2

2 mol XCl = 1.75 mol Cl2

Step 4: Use the ideal gas law to calculate the volume occupied by the gases.

The ideal gas law is given by: P V =nRT where P= Pressure V= Volume n

= Number of moles R= Ideal gas constant T= Temperature

Since the pressure, temperature, and gas constant are constant:

V=nRT

Step 5: Calculate the volume using the number of moles of the gases obtained

in step 3. For Xgas:

VX=3.50 mol ×0.0821 L ·atm/mol ·K×298 K

For Cl2gas :VCl2=1.75 mol×0.0821 L·atm/mol·K×298 K

Step 6: Calculate the total volume occupied by the gases.

Vtotal =VX+VCl2

Question 26

Question

A mixture of methane (CH4) and oxygen (O2) is burned in a constant-volume

container. If 4.50 L of methane at STP is reacted with excess oxygen gas, what

volume of CO2gas is produced?

Solution

Step 1: Write the balanced chemical equation for the combustion of methane

(CH4) in oxygen (O2).

CH4(g) + 2O2(g)→CO2(g) + 2H2O(l)

Step 2: Determine the moles of methane used in the reaction. Given that

the volume of methane is 4.50 L at STP, we can use the formula n=P V

RT where

P= 1 atm, V= 4.50 L, R= 0.0821 L ·atm/K ·mol, and T= 273 K.

nCH4=(1 atm)(4.50 L)

(0.0821 L ·atm/K ·mol)(273 K)

Step 3: Calculate the moles of CO2produced using the mole ratio from the

balanced chemical equation. From the balanced equation, the mole ratio of CH4

to CO2is 1:1.

nCO2=nC H4

Step 4: Convert the moles of CO2to volume using the ideal gas law. Since

the volume at STP is needed, we use the conditions P= 1 atm and T= 273 K.

VCO2=nC O2RT

Step 5: Substitute the values into the formula to find the volume of CO2.

VCO2= (nC O2)(0.0821 L ·atm/K ·mol)(273 K)

Question 27

Question

A sample of solid iron (III) oxide, Fe2O3, is heated strongly in a stream of hy-

drogen gas to form solid iron and water vapor. If 10.0 g of iron (III) oxide is

reacted with an excess of hydrogen gas and yields 5.0 g of iron, what is the the-

oretical yield of water vapor in grams? (Molar masses: Fe2O3= 159.69 g/mol,

Fe = 55.85 g/mol, H2O= 18.02 g/mol)

Solution

Step 1: Calculate the moles of iron produced. Given that the molar mass of iron

is 55.85 g/mol, we can calculate the number of moles of iron produced using the

formula:

moles of Fe = mass of Fe

molar mass of Fe

moles of Fe = 5.0 g

55.85 g/mol = 0.0893 mol

Step 2: Determine the limiting reactant. Next, we need to determine which

reactant is limiting. We will do this by finding the moles of water produced

from the reactants.

Step 2a: Calculate the moles of Fe2O3. Using the given molar mass of

Fe2O3= 159.69 g/mol, we can calculate the moles of Fe2O3as follows:

moles of Fe2O3=mass of Fe2O3

molar mass of Fe2O3

moles of Fe2O3=10.0 g

159.69 g/mol = 0.0626 mol

Step 2b: Use the balanced chemical equation to find the ratio of moles. The

balanced chemical equation is as follows:

Fe2O3+ 3H2→2Fe + 3H2O

The stoichiometry of the reaction tells us that 1 mole of Fe2O3produces 2 moles

of iron and 3 moles of water. Therefore, for the given reaction, 0.0626 moles of

Fe2O3will theoretically produce:

0.0626 mol ×3 mol H2O

1 mol Fe2O3

= 0.1878 mol H2O

Step 3: Calculate the theoretical yield of water vapor. Finally, we use the

calculated moles of water to find the theoretical yield of water vapor in grams

using the molar mass of water:

mass of H2O = moles of H2O×molar mass of H2O

mass of H2O=0.1878 mol ×18.02 g

mol = 3.38 g

Therefore, the theoretical yield of water vapor is 3.38 g.

Question 28

Question

A gaseous compound contains only carbon and hydrogen. When 1.00 g of

this compound is completely burned in excess oxygen, 2.93 g of carbon dioxide

and 0.608 g of water are produced. Determine the empirical formula of the

compound.

Solution

Step 1: Calculate the moles of carbon dioxide and water produced. Given: Mass

of carbon dioxide produced = 2.93 g Mass of water produced = 0.608 g

Calculate the moles of carbon dioxide:

Moles of CO2=Mass of CO2

Molar mass of CO2

=2.93 g

44.01 g/mol = 0.0666 mol

Calculate the moles of water:

Moles of H2O = Mass of H2O

Molar mass of H2O=0.608 g

18.015 g/mol = 0.0337 mol

Step 2: Determine the moles of carbon and hydrogen in the compound.

From the balanced chemical equation for the combustion reaction: 1 mol of

compound produces 1 mol of C dioxide and 1 mol of H2O.

Therefore:

Moles of C = Moles of CO2= 0.0666 mol

Moles of H = Moles of H2O=0.0337 mol

Step 3: Determine the molar ratio of carbon to hydrogen.

Molar ratio of C to H = Moles of C

Moles of H =0.0666 mol

0.0337 mol ≈1.97

Step 4: Determine the empirical formula of the compound. The molar ratio

of carbon to hydrogen is approximately 1.97, which can be approximated to 2.

This suggests that the empirical formula of the compound is C2H4.

Question 29

Question

A gaseous compound containing only carbon and hydrogen is burned in excess

oxygen. If 0.100 moles of the compound produces 40.0 grams of carbon dioxide

and 18.0 grams of water, determine the empirical formula of the compound.

Solution

Step 1: Determine the moles of carbon dioxide and water produced. Given:

Mass of carbon dioxide produced = 40.0 g Molar mass of CO2 = 44.01 g/mol

Moles of CO2 = 40.0

44.01 mol ≈0.909 mol

Mass of water produced = 18.0 g Molar mass of H2O = 18.02 g/mol Moles

of H2O = 18.0

18.02 mol ≈0.999 mol

Step 2: Write the balanced chemical equation. Let the gaseous compound

be represented by CxHy. The balanced chemical equation for the combustion

will be: CH + aO



bCO + cHO

Step 3: Determine the moles of carbon and hydrogen present in the com-

pound. Using the moles of CO2 and H2O as a reference: From CO2: 1 mol C

in CO2 →1 mol C in CH From H2O: 2 mol H in H2O →1 mol H in CH

Moles of C in CH = Moles of CO2 = 0.909 mol Moles of H in CH = 2 ×

Moles of H2O = 2 ×0.999 mol = 1.998 mol

Step 4: Determine the empirical formula. Calculate the ratio of moles of C

to H: C:H = 0.909

1.998 = 0.455 : 1

Since the ratio is approximately 0.5 : 1, the empirical formula of the com-

pound is CH.

Question 30

Question

A mixture of 2.00 L of methane gas (CH4) at STP and 3.00 L of oxygen gas

(O2) at the same temperature and pressure undergoes a combustion reaction.

Calculate the volume of carbon dioxide (CO2) gas produced after the reaction

is complete.

Solution

Step 1: Write the balanced chemical equation for the combustion of methane:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

Step 2: Determine the limiting reactant. First, calculate the number of

moles of each gas using the ideal gas law equation P V =nRT :

For methane:

nCH4=P V

RT =(1 atm)(2.00 L)

(0.0821 atm ·L/mol ·K)(273 K)≈0.083 mol

For oxygen:

nO2=P V

RT =(1 atm)(3.00 L)

(0.0821 atm ·L/mol ·K)(273 K)≈0.1245 mol

Step 3: Use stoichiometry to determine the limiting reactant. Since 1 mole

of methane reacts with 2 moles of oxygen, we can compare the moles of oxygen

with respect to methane:

nO2

2=0.1245

2= 0.06225 mol

Oxygen is the limiting reactant because it will be completely consumed when

all of the methane reacts.

Step 4: Calculate the volume of carbon dioxide produced using the ideal gas

law: Since 1 mole of methane produces 1 mole of carbon dioxide, the number

of moles of CO2produced is also 0.083 mol.

VCO2=nRT

P=(0.083 mol)(0.0821 atm ·L/mol ·K)(273 K)

1atm ≈1.89 L

Therefore, the volume of carbon dioxide gas produced after the reaction is

complete is approximately 1.89 L.

Question 31

Question

When 5.00 liters of oxygen gas at STP react with excess carbon monoxide gas,

how many liters of carbon dioxide gas are produced at STP?

Solution

Step 1: Write down the balanced chemical equation for the reaction:

2CO(g) + O2(g)→2CO2(g)

Step 2: Determine the moles of oxygen gas involved in the reaction. Given

that the volume of oxygen gas is 5.00 L at STP, we can use the molar volume

of a gas at STP to find the number of moles of oxygen:

Molar volume at STP = 22.4 L/mol

Moles of O2=5.00 L

22.4 L/mol = 0.223 mol

Step 3: Use the stoichiometry of the balanced chemical equation to deter-

mine the volume of carbon dioxide gas produced. From the balanced chemical

equation, we can see that 1 mole of O2 reacts with 2 moles of CO2. Therefore,

the number of moles of CO2 produced is twice the number of moles of O2 used:

Moles of CO2= 2 ×Moles of O2= 2 ×0.223 mol = 0.446 mol

Step 4: Calculate the volume of carbon dioxide gas produced at STP. Using

the molar volume of a gas at STP, we can find the volume of carbon dioxide gas

produced:

Volume of CO2= Moles of CO2×Molar volume at STP

Volume of CO2= 0.446 mol ×22.4 L/mol = 9.99 L

Therefore, when 5.00 liters of oxygen gas at STP react with excess carbon

monoxide gas, 9.99 liters of carbon dioxide gas are produced at STP.

Question 32

Question

A mixture of propane gas (C3H8) and oxygen gas (O2) is combusted in a gas

BBQ. If 5.00 L of propane gas at STP (standard temperature and pressure) is

burned in the presence of excess oxygen, what volume of oxygen gas at STP is

needed for complete combustion?

Solution

Step 1: Write the balanced chemical equation for the combustion of propane:

C3H8+ 5O2→3CO2+ 4H2O

Step 2: Calculate the moles of propane: Given the volume of propane gas is

5.00 L at STP, we can use the ideal gas law to calculate the number of moles:

n=P V

n=(1.00 atm)(5.00 L)

0.0821 atm ·L/mol ·K·(273 K)

n= 0.194 mol

Step 3: Use the stoichiometry of the balanced equation to find the volume

of oxygen gas required: 1 mole of propane requires 5 moles of oxygen.

Volume of oxygen gas = 5 ×(Volume of propane gas) = 5 ×5.00 L = 25.0 L

Therefore, 25.0 L of oxygen gas at STP is needed for complete combustion.

Question 33

Question

A reaction takes place between solid aluminum and hydrochloric acid to produce

aluminum chloride and hydrogen gas. If 25.0 grams of aluminum is reacted with

excess hydrochloric acid, how many liters of hydrogen gas at STP are produced?

Solution

Step 1: Write the balanced chemical equation for the reaction. The balanced

chemical equation for the reaction between aluminum and hydrochloric acid is:

2Al + 6HCl →2AlCl3+ 3H2

Step 2: Calculate the molar mass of aluminum. The molar mass of aluminum

(Al) is approximately 26.98 g/mol.

Step 3: Calculate the number of moles of aluminum used.

Moles of Al = Mass

Molar mass =25.0 g

26.98 g/mol ≈0.927 mol

Step 4: Determine the limiting reactant. From the balanced equation, 2

moles of aluminum produce 3 moles of hydrogen gas. Therefore, 0.927 moles of

aluminum will produce:

2×0.927 mol ≈1.390 mol of H2

Step 5: Calculate the volume of hydrogen gas at STP. At STP, 1 mole of

any gas occupies approximately 22.4 L. Therefore, 1.390 moles of hydrogen gas

will occupy:

1.390 mol ×22.4 L/mol = 31.3 L

Final Answer: The reaction of 25.0 grams of aluminum with excess hy-

drochloric acid will produce 31.3 liters of hydrogen gas at STP.

Question 34

Question

A gaseous compound containing only carbon and hydrogen is combusted in

excess oxygen gas. If 3.00 g of the compound produces 8.64 g of carbon dioxide

and 3.52 g of water, what is the empirical formula of the compound?

Solution

Step 1: Calculate the moles of carbon dioxide produced. Given that the molar

mass of carbon dioxide is 44.01 g/mol, we can calculate the moles of carbon

dioxide produced:

moles of CO2=8.64 g

44.01 g/mol = 0.196 mol

Step 2: Calculate the moles of water produced. Given that the molar mass

of water is 18.02 g/mol, we can calculate the moles of water produced:

moles of water = 3.52 g

18.02 g/mol = 0.195 mol

Step 3: Calculate the moles of carbon in the compound. The moles of carbon

is represented by the moles of carbon dioxide produced:

moles of carbon = 0.196 mol

Step 4: Calculate the moles of hydrogen in the compound. Since water is

composed of hydrogen and oxygen, we need to calculate the moles of hydrogen:

moles of H= 2 ×moles of water = 2 ×0.195 mol = 0.390 mol

Step 5: Determine the empirical formula. To find the empirical formula, we

need to determine the ratio of moles of carbon to moles of hydrogen. In this

case, it is a 1:2 ratio, so the empirical formula of the compound is CH2.

Question 35

Question

A reaction takes place between magnesium and hydrochloric acid according to

the following balanced equation:

Mg + 2HCl →MgCl2+ H2

If 3.00 moles of magnesium react with an excess of hydrochloric acid, how

many moles of hydrogen gas are produced?

Step 4: Use the partial pressures to find the moles of each gas.

Moles of CO = 3y×12.39x

0.0821 ×298

Moles of O2=2y×12.39x

0.0821 ×298

Step 5: Set up a system of equations using the molar ratio.

moles of CO

moles of O2

3y×12.39x

0.0821 ×298∇ · 2y×12.39x

0.0821 ×298 =3

3 = 3

Step 6: There seems to be a mistake in the calculations. Let’s reevaluate

the partial pressures and moles of each gas. Reevaluating, we find:

Moles of CO = 3y×12.39x∇ · (0.0821 ×298)

Moles of O2= 2y×12.39x∇ · (0.0821 ×298)

Step 7: Now we can solve the system of equations to find the partial pressures

of each gas.

3y×12.39x

0.0821 ×298 = 3

y=3×0.0821 ×298

3×12.39

y= 6.32 atm

Therefore, the partial pressure of carbon monoxide is 3 ×6.32 = 18.96 atm

and the partial pressure of oxygen is 2 ×6.32 = 12.64 atm.

Question 2

Question

When 10.0 L of butane gas (C4H10) is combusted with excess oxygen gas, how

many liters of carbon dioxide gas will be produced at STP (Standard Temper-

ature and Pressure)? Assume that the combustion of butane is complete.

Solution

Step 1: Write the balanced chemical equation for the combustion of butane:

2C4H10(g) + 13O2(g)→8CO2(g) + 10H2O(g)

Step 2: Calculate the number of moles of butane gas (using ideal gas law

P V =nRT ): Given: Volume of butane gas, V1= 10.0 L Temperature, T=

273 K (STP) Pressure, P= 1 atm Gas constant, R= 0.0821 L ·atm/mol ·K

Using the ideal gas law:

n1=P V

RT =(1 atm)(10.0 L)

(0.0821 L ·atm/mol ·K)(273 K)

n1=10.0

22.37 = 0.447 moles

Step 3: Use the mole ratio from the balanced chemical equation to find

the number of moles of carbon dioxide produced: From the balanced chemical

equation, 2 moles of C4H10 produce 8 moles of CO2. So, for 0.447 moles of

C4H10, the number of moles of CO2produced will be:

Moles of CO2=0.447 moles C4H10 ×8 moles CO2

2 moles C4H10

= 1.79 moles

Step 4: Calculate the volume of carbon dioxide gas at STP: Given that 1

mole of any gas occupies 22.4 L at STP,

V olume = 1.79 moles ×22.4 L/mole = 40.1 L

Therefore, 40.1 liters of carbon dioxide gas will be produced when 10.0 liters

of butane gas is combusted at STP.

Question 3

Question

A reaction occurs between hydrogen gas (H2) and nitrogen gas (N2) according

to the following balanced chemical equation:

N2(g) + 3H2(g)→2NH3(g)

If 5.00 L of nitrogen gas reacts with excess hydrogen gas at STP (standard

temperature and pressure), what volume of ammonia gas is produced?

Solution

Step 1: Calculate the number of moles of nitrogen gas (N2) using the ideal gas

law:

P V =nRT

Since the reaction occurs at STP, the pressure (P) is 1 atm and the tempera-

ture (T) is 273 K. The volume (V) is 5.00 L. The gas constant (R) is 0.0821

L·atm/mol·K.

n=P V

RT =(1.00 atm)(5.00 L)

(0.0821 L ·atm/mol ·K)(273 K)

n=5.00

22.3743 = 0.2234 moles of N2

Step 2: Determine the limiting reactant and the excess reactant.



From the balanced chemical equation, 1 mol of N2reacts with 3 mol of

H2.



Since we have 0.2234 mol of N2, this corresponds to:

0.2234 mol N2×3 mol H2

1 mol N2

= 0.6702 mol H2



If we had an excess of hydrogen gas, the actual number of moles available

for the reaction would be the limiting reactant amount of moles (0.2234

mol). Anything beyond that amount would be in excess.

Step 3: Calculate the volume of ammonia gas (NH3) produced using the

mole ratio from the balanced equation:



From the balanced equation, 1 mol of N2produces 2 mol of NH3.



Hence, 0.2234 mol of N2will produce:

0.2234 mol N2×2 mol NH3

1 mol N2

= 0.4469 mol NH3



Now, we can use the ideal gas law to find the volume of ammonia gas at

STP:

V=nRT

P=(0.4469 mol)(0.0821 L ·atm/mol ·K)(273 K)

1 atm

V=10.8109

1= 10.81 L NH3

Therefore, the volume of ammonia gas produced when 5.00 L of nitrogen gas

reacts with excess hydrogen gas at STP is 10.81 L.

Question 4

Question

A sample of magnesium metal reacts with excess hydrochloric acid to produce

hydrogen gas and magnesium chloride. If 3.00 grams of magnesium completely

reacts, how many liters of hydrogen gas are produced at STP? (Assume the

molar volume of a gas at STP is 22.4 L/mol)

Solution

Step 1: Write the balanced chemical equation for the reaction between magne-

sium and hydrochloric acid. The balanced chemical equation is:

Mg + 2HCl →MgCl2+ H2

Step 2: Calculate the number of moles of magnesium reacting. Given: mass

of magnesium = 3.00 grams Molar mass of magnesium = 24.31 g/mol

Number of moles of magnesium = 3.00 g

24.31 g/mol = 0.1236 mol

Step 3: Use the stoichiometry of the balanced chemical equation to determine

the moles of hydrogen gas produced. From the balanced equation, 1 mole of

magnesium produces 1 mole of hydrogen gas. Number of moles of hydrogen gas

= 0.1236 mol

Step 4: Calculate the volume of hydrogen gas at STP. Given: Molar volume

of gas at STP = 22.4 L/mol

Volume of hydrogen gas = 0.1236 mol ×22.4 L/mol = 2.77 L

Therefore, 2.77 liters of hydrogen gas are produced at STP when 3.00 grams

of magnesium completely reacts.

Question 5

Question

A gaseous compound containing only carbon and hydrogen is burned in oxygen

gas. The products of the combustion reaction are carbon dioxide and water

vapor. If 1.00 g of the compound produces 2.88 g of carbon dioxide and 1.12 g

of water vapor, what is the empirical formula of the compound?

Solution

Step 1: Calculate the moles of carbon dioxide produced. - The molar mass of

carbon dioxide (CO2) is approximately 44.01 g/mol. - Using the given mass of

carbon dioxide:

moles of CO2=2.88 g

44.01 g/mol = 0.06544 mol

Step 2: Calculate the moles of water vapor produced. - The molar mass of

water (H2O) is approximately 18.02 g/mol. - Using the given mass of water:

moles of H2O = 1.12 g

18.02 g/mol = 0.06215 mol

Step 3: Determine the moles of carbon and hydrogen in the compound. -

In the combustion reaction, carbon from the compound forms carbon dioxide,

while hydrogen forms water vapor. - Assume the compound contains xmoles

of carbon and ymoles of hydrogen. - From the moles of products produced:

x= 0.06544 mol of C, y = 0.12430 mol of H

Step 4: Calculate the ratio of moles of carbon and hydrogen. - To find the

simplest whole number ratio of carbon to hydrogen, we need to divide by the

smallest value of xor y, which is 0.06544 mol.

0.06544 = 1,y

0.06544 ≈1.90

Step 5: Find the empirical formula. - Since the ratio is nearly 2, we multiply

both xand yby 2:

2C: 2H⇒C2H2

Step 6: The empirical formula of the compound is C2H2.

Question 6

Question

A gaseous compound containing only carbon and hydrogen was decomposed

completely through a series of reactions. The volume of carbon dioxide gas

produced was found to be 11.2 L at 298 K and 1 atm. If the gas was originally

2.40 g and had a density of 1.964 g/L at the same temperature and pressure,

determine the molecular formula of the compound.

(Molar volume of a gas at STP = 22.4 L/mol)

Solution

Step 1: Find the molar mass of the compound



Given mass of gas, m= 2.40 g



Molar volume of a gas at STP, Vmolar = 22.4 L/mol



Density of the gas, ρ= 1.964 g/L

Step 2: Calculate the molar mass of the compound

Molar mass = mass

number of moles =ρ×Vmolar

Volume of gas

Molar mass = 1.964 g/L ×22.4 L/mol

2.40 g = 18.27 g/mol

Step 3: Determine the empirical formula of the compound



The empirical formula is the simplest ratio of the elements in the com-

pound

Step 4: Determine the molecular formula of the compound



Given molar mass of compound, Mcompound = 18.27 g/mol

Molecular formula = n×Empirical formula

n=Molecular formula molar mass

Empirical formula molar mass =18.27

Empirical formula molar mass

Since the molecular formula of the compound is twice the empirical formula:

n= 2

Therefore, the molecular formula of the compound is twice the empirical

formula.

Question 7

Question

A sample of zinc metal reacts with hydrochloric acid to produce zinc chloride

and hydrogen gas. If 20.0 grams of zinc reacts with an excess of hydrochloric

acid, how many liters of hydrogen gas are produced at STP?

Given: Molar volume of gas at STP = 22.4 L/mol

Solution

Step 1: Write the balanced chemical equation for the reaction.

Zn + 2HCl →ZnCl2+ H2

Step 2: Calculate the number of moles of zinc involved in the reaction.

Molar mass of Zn = 65.38 g/mol

Moles of Zn = 20.0 g

65.38 g/mol

= 0.306 mol

Step 3: Determine the number of moles of hydrogen gas produced using the

mole ratio from the balanced chemical equation.

Moles of H2= 0.306 mol ×1 mol H2

1 mol Zn

= 0.306 mol

Step 4: Calculate the volume of hydrogen gas produced at STP.

Volume of H2= 0.306 mol ×22.4 L/mol

= 6.86 L

Answer: The volume of hydrogen gas produced at STP is 6.86 liters.

Question 8

Question

A gas mixture contains 2.0 moles of oxygen gas (O2) and 3.0 moles of nitrogen

gas (N2). If the temperature and pressure are held constant, what is the total

volume of the gas mixture at standard temperature and pressure (STP)? Assume

ideal gas behavior.

Solution

Step 1: Write the balanced chemical equation for the reaction of oxygen and

nitrogen gases. Step 2: Determine the mole ratio of oxygen to nitrogen in the

reaction. Step 3: Use the ideal gas law to calculate the total volume of the gas

mixture at STP.

Step 1: Write the balanced chemical equation for the reaction of oxygen and

nitrogen gases. The balanced chemical equation for the reaction of oxygen and

nitrogen gases is:

2O2(g) + N2(g)→2O2N2(g)

Step 2: Determine the mole ratio of oxygen to nitrogen in the reaction. From

the balanced chemical equation, the mole ratio of oxygen to nitrogen is 2:1.

Step 3: Use the ideal gas law to calculate the total volume of the gas mixture

at STP. At STP (standard temperature and pressure), the conditions are: -

Temperature = 273 K - Pressure = 1 atm - Volume of 1 mole of gas at STP =

22.4 L

Since the mole ratio of oxygen to nitrogen is 2:1, we can find the moles of

oxygen and nitrogen in the mixture: - Moles of oxygen: 2.0 moles - Moles of

nitrogen: 3.0 moles

Now, we can calculate the total volume of the gas mixture using the ideal

gas law:

P V =nRT

V=nRT

V=(2.0+3.0) ×0.0821 ×273

V=5.0×0.0821 ×273

V=112.27

V≈112.27 L

Therefore, the total volume of the gas mixture at STP is approximately

112.27 L.

Question 9

Question

In a reaction involving the combustion of butane gas (C4H10), 12 moles of

butane gas react with excess oxygen gas to produce carbon dioxide and water

vapor. Calculate the volume of carbon dioxide produced at STP.

Solution

Step 1: Write the balanced chemical equation for the combustion of butane:

2C4H10 + 13O2→8CO2+ 10H2O

Step 2: Determine the molar ratio between butane and carbon dioxide from

the balanced equation: 1 mole of C4H10 reacts to produce 8 moles of CO2

Step 3: Calculate the number of moles of carbon dioxide produced when 12

moles of butane react:

12 moles of C4H10 ×8 moles of CO2

2 moles of C4H10

= 48 moles of CO2

Step 4: Convert moles of carbon dioxide to volume at STP (standard tem-

perature and pressure, which is 0

C and 1 atm pressure):

48 moles of CO2×22.4 L/mol = 1075.2 L of CO2

Therefore, 1075.2 L of carbon dioxide is produced during the combustion of

12 moles of butane at STP.

Question 10

Question

A 4.00 L flask contains gas at a pressure of 3.00 atm and a temperature of 25

This gas is then transferred to a 2.00 L flask at 25

C. What is the new pressure

inside the 2.00 L flask?

(Hint: Remember to use the ideal gas law, P V =nRT , where Pis pressure,

Vis volume, nis the number of moles of gas, Ris the gas constant, and Tis

temperature in Kelvin.)

Solution

Step 1: Convert the temperature from Celsius to Kelvin. Given that the

temperature is 25

C, we can convert it to Kelvin using the formula T(K) =

T(C) + 273.15. Therefore, T(K) = 25 + 273.15 = 298.15 K.

Step 2: Calculate the number of moles of gas using the ideal gas law, P V =

nRT . For the initial condition in the 4.00 L flask: P1= 3.00 atm, V1= 4.00 L,

T1= 298.15 K.

n=P1V1

RT1=(3.00 atm)(4.00 L)

0.0821 L atm/mol K·298.15 K n≈0.486 mol.

Step 3: Apply the ideal gas law to find the new pressure in the 2.00 L flask.

Since moles of gas remain constant, we have: P1V1=P2V2.

P2=P1V1

V2=(3.00 atm)(4.00 L)

2.00 L .P2= 6.00 atm.

Therefore, the new pressure inside the 2.00 L flask is 6.00 atm.

Question 11

Question

A gaseous compound containing only carbon and hydrogen was burned in excess

oxygen. The products were carbon dioxide and water vapor. If 0.550 grams of

the compound produced 1.560 grams of carbon dioxide and 0.640 grams of water

vapor, determine the empirical formula of the compound.

Solution

Step 1: Find the moles of each product produced. Given: - Mass of compound

burned = 0.550 g - Mass of carbon dioxide produced = 1.560 g - Mass of water

vapor produced = 0.640 g

Calculate the moles of carbon dioxide produced:

Moles of CO2=1.560 g

44.01 g/mol = 0.0355 mol

Calculate the moles of water vapor produced:

Moles of H2O=0.640 g

18.02 g/mol = 0.0355 mol

Step 2: Find the moles of carbon and hydrogen in the compound. From the

balanced chemical equation for the combustion of the compound:

CaHb+ O2→CO2+ H2O

we see that 1 mole of carbon compound produces 1 mole of CO2and 1 mole of

H2O.

Since the moles of carbon dioxide produced is equal to the moles of water

vapor produced, the moles of carbon produced is also equal to the moles of

hydrogen produced.

Step 3: Find the empirical formula of the compound. Calculate the molar

mass of each product: - CO2: 12.01 + 2(16.00) = 44.01 g/mol - H2O: 2(1.01) +

16.00 = 18.02 g/mol

The molar mass of CO2consists of 1 mole of carbon and 2 moles of oxygen,

while the molar mass of H2Oconsists of 2 moles of hydrogen and 1 mole of

oxygen.

Since carbon and hydrogen are the only elements in the compound, the molar

mass of the compound is approximately:

(12.01a+ 1.01b) g/mol

From the moles of carbon dioxide and water vapor produced, we know that:

0.0355 = a

12.01 =b

1.01

Solving these equations simultaneously, we find: a= 1 and b= 1

Therefore, the empirical formula of the compound is CH.

Question 12

Question

In the combustion of propane (C3H8) with excess oxygen, propane is converted

to carbon dioxide and water vapor according to the following balanced equation:

C3H8+ 5O2→3CO2+ 4H2O

If 20.0 g of propane is burned, how many grams of water vapor are produced?

Solution

Step 1: Calculate the number of moles of propane burned. Given: Mass of

propane, C3H8: 20.0 g Molar mass of C3H8: 3(12.01 g/mol) + 8(1.008 g/mol)

= 44.11 g/mol

Number of moles of C3H8=20.0 g

44.11 g/mol = 0.453 mol

Step 2: Determine the limiting reactant. Using the balanced equation, we

see that 1 mole of C3H8produces 4 moles of H2O. Therefore, 0.453 moles of

C3H8will produce: 0.453 mol C3H8×4 mol H2O

1 mol C3H8

= 1.81 mol H2O

Step 3: Calculate the mass of water vapor produced. Molar mass of H2O:

2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol

Mass of H2Oproduced = 1.81 mol ×18.02 g/mol = 32.6 g

Therefore, 32.6 grams of water vapor are produced when 20.0 grams of

propane is burned.

Question 13

Question

A gaseous compound containing carbon, hydrogen, and oxygen was burned in

excess oxygen gas. The combustion of 0.300 g of this compound produced 0.882

g of carbon dioxide and 0.506 g of water. Determine the empirical formula of

the compound.

Solution

Step 1: Calculate moles of carbon dioxide produced. The molar mass of carbon

dioxide (CO2) is 44.01 g/mol. Therefore,

moles of CO2=0.882 g

44.01 g/mol = 0.02 mol

Step 2: Calculate moles of water produced. The molar mass of water (H2O)

is 18.015 g/mol. Therefore,

moles of H2O = 0.506 g

18.015 g/mol = 0.028 mol

Step 3: Determine the moles of carbon, hydrogen, and oxygen in the com-

pound. Let the moles of carbon, hydrogen, and oxygen in the compound be

represented by x, y, and z respectively. From the chemical equation of the

combustion reaction, we have the following relationships:

1 mole of C produces 1 mole of CO2.

1 mole of H2produces 1/2 moles of H2O.

Using the given information, we can set up the following equations:

x= 0.02 mol of C

y= 2 ×0.028 mol of H

2x+y=z

Step 4: Calculate the molar ratios. The simplest way to proceed is to choose

a common multiplier that will make all the moles whole numbers. By inspection,

we can see that multiplying all values by 50 will give whole numbers:

x= 1 mol of C

y= 2 mol of H

z= 100 mol of O

Step 5: Determine the empirical formula. The empirical formula of the

compound is C1H2O100. Simplifying, we find that the empirical formula is

CHO50.

Question 14

Question

A gaseous compound containing only carbon and hydrogen is burned in excess

oxygen. It is found that 1.00 L of the gas at 0.980 atm and 27.0

C gives 2.17 g

CO2at the same temperature and pressure. What is the molecular formula of

the compound?

(Molar masses: C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol)

Solution

Step 1: Calculate the moles of CO2produced. Given mass of CO2= 2.17 g

Molar mass of CO2= 44.01 g/mol (from the atomic masses of C and O) Number

of moles of CO2=2.17 g

44.01 g/mol = 0.0493 mol

Step 2: Calculate the moles of O2consumed. From the balanced chemi-

cal equation for the combustion of the gaseous compound: 1 mole of gaseous

compound + 3

2moles of O2→2 moles of CO2+ x moles of H2O

Since the gaseous compound contains only carbon and hydrogen, the only

source of carbon in the products is from CO2. Therefore: 1 mole of gaseous

compound →2 moles of CO2

So, moles of O2consumed = 2 ×moles of CO2= 2 ×0.0493 = 0.0986 mol

Step 3: Calculate the moles of gaseous compound. Since the gaseous com-

pound is burned in excess oxygen, all carbon and hydrogen in the compound

end up in the products. Therefore, the moles of gaseous compound consumed in

the reaction is equal to the moles of O2consumed: Moles of gaseous compound

= 0.0986 mol

Step 4: Calculate the molar mass of the gaseous compound. Given moles of

gaseous compound = 0.0986 mol Given volume of gaseous compound = 1.00 L

Given pressure of gaseous compound = 0.980 atm Given temperature of gaseous

compound = 27.0

C = 300.0 K

Using the ideal gas law, P V =nRT , we can rearrange to find molar mass:

M=mRT

P V

Substitute the given values: M=(0.0986 mol)(0.0821 L·atm/mol/K)(300.0 K)

(0.980 atm)(1.00 L)

M=2.4480

0.980 = 2.5 g/mol

Step 5: Determine the empirical formula of the compound. Given the masses

of carbon and hydrogen in the compound, we can find the mole ratio of C to

H in the compound: Molar mass of C = 12.01 g/mol Molar mass of H = 1.01

g/mol

Assume 1 mole of the compound contains x moles of carbon and y moles of

hydrogen: 12.01x + 1.01y = 2.5

Solving for the smallest whole number ratio of x to y gives the empirical

formula of the compound.

Step 6: Calculate the molecular formula of the compound. Use the molar

mass of the compound to determine the molecular formula. For example, if the

empirical formula is CH, and the molar mass is 50 g/mol, then the molecular

formula would be C2H2.

Question 15

Question

A gaseous hydrocarbon reacts with oxygen according to the following balanced

chemical equation:

CnHm+ (n+m

4)O2→nCO2+m

2H2O

If 5.00 g of the hydrocarbon reacts with 50.0 g of oxygen and produces 13.2 g of

carbon dioxide, determine the empirical formula of the hydrocarbon. (Assume

all substances are gases at the conditions of the reaction.)

Solution

Step 1: Convert the masses given to moles. The molar mass of oxygen (O2) is

32.00 g/mol. The molar mass of carbon dioxide (CO2) is 44.01 g/mol. We can

use these molar masses to convert grams to moles. For the hydrocarbon, we do

not know the molar mass, so we’ll denote it as M.

Given: Mass of hydrocarbon = 5.00 g Mass of oxygen = 50.0 g Mass of CO2

produced = 13.2 g

Number of moles of hydrocarbon:

moles of hydrocarbon = 5.00 g

Number of moles of oxygen:

moles of oxygen = 50.0 g

32.00 g/mol

Number of moles of CO2 produced:

moles of CO2 = 13.2 g

44.01 g/mol

Step 2: Use stoichiometry to relate the moles of reactants and products to

find the empirical formula. From the balanced chemical equation, 1 mole of

hydrocarbon reacts with n+m/4 moles of O2to produce nmoles of CO2. This

can be used to set up the following ratio:

moles of hydrocarbon

moles of oxygen =

5.00

50.0

32.00

n+m/4

Similarly, the ratio of moles of hydrocarbon to moles of CO2is:

moles of hydrocarbon

moles of CO2 =

5.00

13.2

44.01

Solving these two equations simultaneously will allow us to determine the

values of nand m, and hence, the empirical formula of the hydrocarbon.