CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY -
Stoichiometry
Question Bank - Set 2
Liberty University
Question 1
Question
A student wants to determine the percentage composition of a compound con-
taining only carbon, hydrogen, and oxygen. To do this, the student combusts
a 1.50 g sample of the compound in excess oxygen, producing 2.80 g of car-
bon dioxide and 1.15 g of water. Calculate the percentage composition of the
compound.
Solution
Step 1: Calculate the moles of carbon dioxide produced. Given the molar mass
of carbon dioxide (CO2) is 44.01 g/mol, we can calculate the moles of carbon
dioxide produced:
moles of CO2=2.80 g
44.01 g/mol = 0.0636 mol
Step 2: Calculate the moles of water produced. Given the molar mass of
water (H2O) is 18.02 g/mol, we can calculate the moles of water produced:
moles of H2O=1.15 g
18.02 g/mol = 0.0638 mol
Step 3: Determine the moles of carbon, hydrogen, and oxygen in the com-
pound. From the balanced chemical equation for the combustion of the com-
pound, we know that 1 mol of carbon in the compound yields 1 mol of CO2
and 1 mol of hydrogen yields 1
2mol of H2O. Therefore, we have: - Moles of
carbon in the compound = 0.0636 mol - Moles of hydrogen in the compound =
2×0.0638 mol = 0.1276 mol - Moles of oxygen in the compound = Moles of
CO2+ Moles of H2O= 0.0636 mol + 0.0638 mol = 0.1274 mol
Step 4: Calculate the molar mass of the compound. The molar mass of the
compound is the sum of the molar masses of carbon, hydrogen, and oxygen:
Molar mass of compound = 12.01 g/mol+1.01 g/mol+16.00 g/mol = 29.02 g/mol
Step 5: Calculate the percentage composition of the compound. The per-
centage composition of each element in the compound is calculated by dividing
the moles of the element by the total moles of the compound, then multiplying
by 100- Carbon: 0.0636
0.29 ×100% = 21.9% - Hydrogen: 0.1276
0.29 ×100% = 44.0% -
Oxygen: 0.1274
0.29 ×100% = 34.1%
Therefore, the compound is composed of 21.9
Question 2
Question
A chemist has 5.00 moles of sulfuric acid (H2SO4) and wants to completely
neutralize it with calcium hydroxide (Ca(OH)2) to produce calcium sulfate
(CaSO4) and water. Calculate the mass of calcium hydroxide needed for the
reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid and calcium hydroxide.
H2SO4+Ca(OH)2→CaSO4+ 2H2O
Step 2: Determine the molar ratio between H2SO4and Ca(OH)2from the
balanced equation.
From the equation, 1 mole of H2SO4reacts with 1 mole of Ca(OH)2.
Step 3: Calculate the number of moles of Ca(OH)2needed to neutralize
5.00 moles of H2SO4.
Since the molar ratio between H2SO4and Ca(OH)2is 1:1, 5.00 moles of
H2SO4will require 5.00 moles of Ca(OH)2.
Step 4: Determine the molar mass of Ca(OH)2.
Ca : 1 ×40.08 g/mol O: 2 ×16.00 g/mol H: 2 ×1.01 g/mol = 40.08 +
32.00 + 2.02 g/mol = 74.10 g/mol
Step 5: Calculate the mass of Ca(OH)2needed to neutralize the sulfuric
acid.
Mass = Number of moles×Molar mass = 5.00 moles×74.10 g/mol = 370.50
g
Therefore, the chemist will need 370.50 grams of calcium hydroxide to com-
pletely neutralize 5.00 moles of sulfuric acid.
2
Question 3
Question
A reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) pro-
duces potassium sulfate (K2SO4) and water. If 50.0 mL of 0.500 M sulfuric
acid is reacted with excess potassium hydroxide, calculate the volume of 1.00
M potassium hydroxide needed to completely react with the sulfuric acid.
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid and potassium hydroxide.
H2SO4+ 2KOH →K2SO4+ 2H2O
Step 2: Determine the number of moles of sulfuric acid using the given
concentration and volume.
moles of H2SO4= Molarity ×Volume (L)
moles of H2SO4= 0.500 mol/L ×0.0500 L
moles of H2SO4= 0.0250 mol
Step 3: Use the mole ratio from the balanced equation to find the moles of
potassium hydroxide needed. From the equation, 1 mole of H2SO4reacts with
2 moles of KOH. Therefore, moles of KOH = 2 ×moles of H2SO4.
moles of KOH = 2 ×0.0250 mol
moles of KOH = 0.0500 mol
Step 4: Calculate the required volume of 1.00 M potassium hydroxide using
the moles and concentration.
Molarity = moles
volume (L) ⇒volume = moles
Molarity
volume of 1.00 M KOH = 0.0500 mol
1.00 mol/L
volume of 1.00 M KOH = 0.0500 L = 50.0 mL
Therefore, 50.0 mL of 1.00 M potassium hydroxide is needed to completely
react with 50.0 mL of 0.500 M sulfuric acid.
Question 4
Question
A sample of iron(III) chloride, FeCl3, is found to contain 2.50 moles of chlorine
atoms. How many moles of iron atoms are present in the sample?
3
Solution
Step 1: Write the balanced chemical equation for the dissociation of iron(III)
chloride.
FeCl3→Fe3+ + 3Cl−
Step 2: Determine the mole ratio between chlorine atoms and iron atoms
from the balanced chemical equation. For every mole of iron(III) chloride, there
are 1 mole of iron atoms and 3 moles of chlorine atoms. So, the mole ratio
between iron atoms and chlorine atoms is 1:3.
Step 3: Use the mole ratio to find the number of moles of iron atoms. Since
we have 2.50 moles of chlorine atoms, we can set up the following proportion:
2.50 mol Cl
3=xmol Fe
1
Step 4: Solve for the number of moles of iron atoms:
x=2.50 ×1
3= 0.833 mol Fe
Therefore, there are 0.833 moles of iron atoms present in the sample of
iron(III) chloride.
Question 5
Question
A student is performing a reaction in the lab that produces 2 moles of product
F for every 3 moles of reactant R. If the student starts with 4 moles of reactant
R, how many moles of product F will be produced?
Solution
Step 1: Determine the mole ratio between reactant R and product F. The mole
ratio between reactant R and product F is 3 moles of R to 2 moles of F.
Step 2: Calculate the moles of product F produced from 4 moles of reactant
R. Given that there are 4 moles of reactant R, we can set up a proportion to
find the moles of product F produced:
4 moles R
3=xmoles of F
2
Step 3: Solve for x to find the moles of product F. Cross multiplying, we
get:
4×2=3x
8=3x
x=8
3=2
3×8 = 16
3
Thus, 16/3 moles of product F will be produced when 4 moles of reactant R are
used.
4
Question 6
Question
A chemist wishes to produce 250.0 g of iron (III) oxide according to the following
balanced chemical equation:
4F e + 3O2→2F e2O3
If iron is available as iron (III) chloride, FeCl3, how many grams of iron (III)
chloride should be used to produce the desired amount of iron (III) oxide?
Solution
Step 1: Find the molar mass of iron (III) oxide (F e2O3). The molar mass of
F e2O3is calculated by adding the molar mass of iron (Fe) and three times the
molar mass of oxygen (O).
= 2 ×Atomic mass of Fe + 3 ×Atomic mass of O
= 2 ×55.85 g/mol + 3 ×16.00 g/mol
= 159.70 g/mol
Step 2: Calculate the number of moles of iron (III) oxide needed. Given
mass of iron (III) oxide, mF e2O3= 250.0 g. Using the molar mass calculated
above, we can find the number of moles of F e2O3using the formula:
n=m
M
n=250.0 g
159.70 g/mol
n≈1.567 mol
Step 3: Determine the mole ratio between iron (III) chloride and iron (III)
oxide. From the balanced chemical equation, the mole ratio between iron (III)
chloride (FeCl3) and iron (III) oxide (F e2O3) is 4:2. This can be simplified to
a ratio of 2:1.
Step 4: Convert moles of iron (III) oxide to moles of iron (III) chloride. Since
the mole ratio is 2:1, the number of moles of iron (III) chloride needed is half
the number of moles of iron (III) oxide:
nF eCl3=1.567 mol
2
nF eCl3= 0.784 mol
Step 5: Calculate the mass of iron (III) chloride required. Using the molar
mass of iron (III) chloride (F eCl3):
MF eCl3= 162.20 g/mol
5
The mass of iron (III) chloride needed can be calculated using the formula:
mF eCl3=nF eCl3×MF eCl3
mF eCl3= 0.784 mol ×162.20 g/mol
mF eCl3≈127.24 g
Therefore, approximately 127.24 grams of iron (III) chloride should be used
to produce 250.0 grams of iron (III) oxide.
Question 7
Question
A chemist is studying the reaction between silver nitrate (AgNO3) and sodium
chloride (NaCl). If 10.0 grams of silver nitrate reacts with excess sodium chlo-
ride, what is the theoretical yield of silver chloride that can be produced? The
balanced chemical equation for the reaction is:
AgNO3+ NaCl →AgCl + NaNO3
Solution
Step 1: Determine the molar mass of silver nitrate and calculate the number of
moles. The molar mass of AgNO3is:
1Ag + 1N + 3O = 107.87g/mol + 14.01g/mol + 3(16.00g/mol) = 169.87g/mol
The number of moles of silver nitrate is calculated as:
moles = mass
molar mass =10.0g
169.87g/mol = 0.0589 mol
Step 2: Use the balanced chemical equation to determine the stoichiometry
of the reaction. From the balanced chemical equation, we see that 1 mole of
AgNO3produces 1 mole of AgCl.
Step 3: Calculate the theoretical yield of silver chloride. Since the stoichio-
metric ratio is 1:1, the number of moles of AgCl produced will be the same as
the number of moles of AgNO3used.
The mass of silver chloride produced is calculated as:
mass = moles ×molar mass = 0.0589 mol ×143.32 g/mol = 8.46 g
Therefore, the theoretical yield of silver chloride that can be produced is
8.46 grams.
6
Question 8
Question
A student wants to determine the amount of potassium permanganate, KM nO4,
needed to titrate a 0.500 M solution of iron(II) ion, F e2+, in acidic solution.
The balanced chemical equation for the reaction is:
5F e2+ + 8H++MnO−
4→5F e3+ +Mn2+ + 4H2O
If it takes 40.0 mL of the potassium permanganate solution to reach the equiv-
alence point, what mass of KMnO4in grams would be needed?
(Hint: The molar mass of KM nO4is 158.03 g/mol)
Solution
Step 1: Determine the moles of F e2+ in the solution. Given that the volume of
the F e2+ solution is 40.0 mL and its concentration is 0.500 M, we can calculate
the moles of F e2+.
Moles of F e2+ = Volume×Concentration = 40.0×10−3L×0.500 mol/L = 0.0200 mol
Step 2: Use the stoichiometry of the balanced equation to determine the
moles of KMnO4needed. From the balanced chemical equation, we see that 1
mole of MnO−
4reacts with 5 moles of F e2+.
Moles of KMnO4=Moles of F e2+
5=0.0200 mol
5= 0.00400 mol
Step 3: Calculate the mass of KM nO4needed. Using the molar mass of
KMnO4(158.03 g/mol), we can find the mass of KM nO4required.
Mass of KMnO4= Moles×Molar Mass = 0.00400 mol×158.03 g/mol = 0.632 g
Therefore, the mass of KMnO4needed to titrate the F e2+ solution is 0.632
grams.
Question 9
Question
A sample of an unknown compound containing carbon, hydrogen, and oxygen
is analyzed. Combustion of a 0.568 g sample of the compound yields 1.105 g
of carbon dioxide and 0.471 g of water. What is the empirical formula of the
compound?
(Molar masses: C = 12.011 g/mol, H = 1.008 g/mol, O = 15.999 g/mol)
7
Solution
Step 1: Find the moles of carbon and hydrogen in the compound. - The molar
mass of carbon dioxide is 44.01 g/mol, so the moles of carbon dioxide produced
can be calculated as:
moles of CO2=1.105 g
44.01 g/mol = 0.0251 mol
- Since each mole of carbon dioxide contains 1 mole of carbon, the moles of
carbon in the compound is also 0.0251 mol.
- The molar mass of water is 18.015 g/mol, so the moles of water produced
can be calculated as:
moles of H2O=0.471 g
18.015 g/mol = 0.0261 mol
- Since each mole of water contains 2 moles of hydrogen, the moles of hydro-
gen in the compound is 2 * 0.0261 mol = 0.0522 mol.
Step 2: Determine the molar ratios of carbon, hydrogen, and oxygen in the
compound. - The ratio of carbon to hydrogen is 0.0251 mol : 0.0522 mol, which
simplifies to 1 : 2.08.
Step 3: Convert the molar ratios to whole numbers. - The molar ratio of
carbon to hydrogen can be approximated to 1 : 2.
Step 4: Write the empirical formula using the whole-number ratios. - The
empirical formula of the compound is CH2.
Question 10
Question
A sample of aluminum chloride (AlCl3) reacts with excess sodium carbonate
(Na2CO3) to produce aluminum carbonate (Al2(CO3)3) and sodium chloride
(NaCl). If 15.0 grams of aluminum chloride are reacted with 25.0 grams of
sodium carbonate, what mass of aluminum carbonate will be produced?
(Molar masses: AlCl3= 133.34 g/mol, Na2CO3= 105.99 g/mol, Al2(CO3)3=
324.95 g/mol, NaCl = 58.44 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
chloride and sodium carbonate.
2AlCl3+ 3N a2CO3→Al2(CO3)3+ 6N aCl
8
Step 2: Calculate the number of moles of each reactant.
Moles of AlCl3=15.0 g
133.34 g/mol
= 0.1124 mol
Moles of Na2CO3=25.0 g
105.99 g/mol
= 0.2360 mol
Step 3: Determine the limiting reactant by examining the mole ratio between
aluminum chloride and sodium carbonate from the balanced equation.
Moles of AlCl3
2=0.1124 mol
2= 0.0562 mol
Moles of Na2CO3
3=0.2360 mol
3= 0.0787 mol
Since aluminum chloride produces the least amount of product, it is the limiting
reactant.
Step 4: Calculate the theoretical yield of aluminum carbonate using the
limiting reactant.
Moles of Al2(CO3)3= 0.0562 mol ×1 mol Al2(CO3)3
2 mol AlCl3
= 0.0281 mol
Mass of Al2(CO3)3= 0.0281 mol ×324.95 g/mol
= 9.13 g
Answer: The mass of aluminum carbonate produced is 9.13 grams.
Question 11
Question
A reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) pro-
duces sodium sulfate (Na2SO4) and water (H2O).
Given a 1.50 L solution of H2SO4with a concentration of 0.500 M, determine
the volume of 2.00 M NaOH solution needed to completely react with the
sulfuric acid. Assume that the reaction goes to completion and that the volumes
of the solutions are additive.
Solution
Step 1: Write the balanced chemical equation for the reaction between H2SO4
and NaOH.
H2SO4(aq)+2NaOH(aq)→N a2SO4(aq)+2H2O(l)
9
Step 2: Determine the number of moles of H2SO4present in the solution.
Given: Volume of H2SO4solution = 1.50 L Concentration of H2SO4solution
= 0.500 M
Using the formula C=n
V(where C= concentration, n= moles, and V=
volume), we have: n=C×V n = 0.500 mol/L ×1.50 L n= 0.750 moles
Step 3: Use the stoichiometry of the balanced chemical equation to find the
amount of NaOH needed to react with all the H2SO4. From the balanced
chemical equation, we see that 1 mole of H2SO4reacts with 2 moles of NaOH.
Therefore, to react completely with 0.750 moles of H2SO4, we need: 2 ×
0.750 = 1.50 moles of NaOH
Step 4: Calculate the volume of 2.00 M NaOH solution needed. Concen-
tration of NaOH solution = 2.00 M Volume of NaOH solution needed can be
calculated using the formula: V=n
C(where V= volume, n= moles, and C=
concentration)
We have: V=1.50 moles
2.00 mol/L V= 0.75 L
Therefore, 0.75 L of 2.00 M NaOH solution is needed to completely react
with the sulfuric acid.
Question 12
Question
A sample of calcium carbonate, CaCO3, with a mass of 5.00 g, reacts completely
with excess hydrochloric acid solution according to the following balanced chem-
ical equation:
CaCO3(s) + 2HCl(aq)→CaCl2(aq) + CO2(g)+H2O(l)
Calculate the volume of carbon dioxide produced at STP.
Solution
Step 1: Calculate the number of moles of calcium carbonate.
Molar mass of CaCO3= 40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol)
= 100.09 g/mol
Number of moles of CaCO3=Mass
Molar mass
=5.00 g
100.09 g/mol
= 0.04997 mol
Step 2: Use the balanced chemical equation to relate the moles of carbon
dioxide produced to the moles of calcium carbonate. According to the balanced
10
chemical equation, 1 mole of CaCO3produces 1 mole of CO2. Thus, 0.04997
mol of CaCO3will produce 0.04997 mol of CO2.
Step 3: Calculate the volume of carbon dioxide produced at STP. At STP
(Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.
Volume of CO2= Number of moles ×22.4 L/mol
= 0.04997 mol ×22.4 L/mol
= 1.12 L
Therefore, the volume of carbon dioxide produced at STP is 1.12 liters.
Question 13
Question
A chemist is trying to synthesize a compound using the following reaction:
2C6H6+ 15O2→12CO2+ 6H2O
If the chemist has 200 grams of benzene (C6H6) and plenty of oxygen gas (O2),
how many grams of water (H2O) will be produced in theory?
Solution
Step 1: Find the molar mass of benzene (C6H6). The molar mass of carbon
(C) is 12.01 g/mol and the molar mass of hydrogen (H) is 1.008 g/mol. Since
benzene has 6 carbons and 6 hydrogens, the molar mass of benzene is:
6(12.01 g/mol) + 6(1.008 g/mol) = 78.11 g/mol
Step 2: Calculate the number of moles of benzene. Given that the chemist
has 200 grams of benzene:
200 g
78.11 g/mol = 2.563 mol
Step 3: Determine the limiting reagent between benzene and oxygen. From
the balanced chemical equation, we can see that 2 moles of benzene react with
15 moles of oxygen gas. Let’s calculate the number of moles of oxygen gas
needed to react with the given amount of benzene:
2.563 mol C6H6×15 mol O2
2 mol C6H6
= 19.218 mol O2
Step 4: Calculate the mass of water produced. From the balanced chemical
equation, we can see that 2 moles of water are produced for every 15 moles of
oxygen gas. Let’s calculate the mass of water produced:
19.218 mol O2×6 mol H2O
15 mol O2
×18.015 g/mol H2O = 137.09 g H2O
Therefore, in theory, 137.09 grams of water will be produced.
11
Question 14
Question
A compound containing only carbon, hydrogen, and oxygen is burned in oxygen
gas. In one experiment, a 0.500 g sample of the compound produced 0.964 g
of carbon dioxide and 0.396 g of water. Determine the empirical formula of the
compound.
Solution
Step 1: Calculate moles of carbon dioxide produced.
Moles of CO2=Mass of CO2
Molar mass of CO2
=0.964 g
44.01 g/mol = 0.0219 mol
Step 2: Calculate moles of water produced.
Moles of H2O=Mass of H2O
Molar mass of H2O=0.396 g
18.02 g/mol = 0.0220 mol
Step 3: Determine the moles of carbon in the original compound. Since all
the carbon in the compound ends up in the form of CO2, the moles of carbon
in the original compound = moles of CO2produced. Moles of carbon = 0.0219
mol
Step 4: Determine the moles of hydrogen in the original compound. Since
all the hydrogen in the compound ends up in the form of H2O, the moles
of hydrogen in the original compound = moles of H2Oproduced. Moles of
hydrogen = 0.0220 mol
Step 5: Calculate the moles of oxygen in the original compound. The moles
of oxygen in the original compound can be calculated using the fact that the
sum of moles of carbon, hydrogen, and oxygen in the compound is equal to the
total moles of the compound. Total moles of compound = moles of carbon +
moles of hydrogen + moles of oxygen Total moles of compound = 0.500 g /
molar mass of compound
Step 6: Determine the empirical formula of the compound. Now that we have
the moles of carbon, hydrogen, and oxygen in the compound, we can determine
the empirical formula. Divide the moles of each element by the smallest number
of moles obtained:
Carbon: 0.0219 mol/0.0219 mol = 1
Hydrogen: 0.0220 mol/0.0219 mol ≈1
Oxygen: (Total moles of compound −moles of carbon −moles of hydrogen)/0.0219 mol
Thus, the empirical formula of the compound is CH1Ox.
12
Question 15
Question
A reaction between silver nitrate (AgNO3) and sodium phosphate (Na3PO4)
produces silver phosphate precipitate (Ag3PO4) and sodium nitrate (NaNO3).
If 5.00 g of silver nitrate is reacted with an excess of sodium phosphate and 8.00
g of silver phosphate is isolated, what is the percent yield of the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction is:
3AgNO3+ Na3PO4→Ag3PO4+ 3NaNO3
Step 2: Calculate the molar mass of silver nitrate (AgNO3).
Molar mass of AgNO3= molar massAg + molar massN+ 3 ×molar massO
= 107.87 g/mol + 14.01 g/mol + 3 ×16.00 g/mol = 169.87 g/mol
Step 3: Calculate the number of moles of silver nitrate used.
Moles of AgNO3=mass of AgNO3
molar mass of AgNO3
=5.00 g
169.87 g/mol = 0.0294 mol
Step 4: Determine the theoretical yield of silver phosphate. From the bal-
anced chemical equation, it is clear that the molar ratio of AgNO3to Ag3PO4
is 3:1. Therefore, the theoretical yield of Ag3PO4is:
Theoretical yield of Ag3PO4= 3×moles of AgNO3= 3×0.0294 mol = 0.0882 mol
Step 5: Calculate the molar mass of silver phosphate (Ag3PO4).
Molar mass of Ag3PO4= 3 ×molar massAg + molar massP+ 4 ×molar massO
= 3 ×107.87 g/mol + 30.97 g/mol + 4 ×16.00 g/mol = 418.62 g/mol
Step 6: Calculate the actual yield of silver phosphate.
Actual yield of Ag3PO4= 8.00 g
Step 7: Calculate the percent yield of the reaction.
Percent yield = Actual yield
Theoretical yield ×100% = 8.00 g
0.0882 mol ×418.62 g/mol ×100%
= 44.6%
Therefore, the percent yield of the reaction is 44.6
13
Question 16
Question
A chemical reaction between sulfuric acid (H2SO4) and potassium hydroxide
(KOH) produces potassium sulfate and water. If 25.0 mL of 2.00 M sulfuric
acid reacts with excess potassium hydroxide, what mass of potassium sulfate is
produced?
Solution
Step 1: Write the balanced chemical equation for the reaction.
H2SO4+ 2KOH →K2SO4+ 2H2O
Step 2: Determine the number of moles of sulfuric acid (H2SO4) used.
moles of H2SO4= volume ×molarity = 0.025 L ×2.00 mol/L = 0.050 mol
Step 3: Use the mole ratio from the balanced equation to find the number
of moles of potassium sulfate (K2SO4) produced.
moles of K2SO4=0.050 mol H2SO4×1 mol K2SO4
1 mol H2SO4
= 0.050 mol K2SO4
Step 4: Calculate the mass of potassium sulfate produced using the molar
mass of K2SO4.
m= moles ×molar mass = 0.050 mol ×174.259 g/mol = 8.71 g
Therefore, 8.71 g of potassium sulfate is produced when 25.0 mL of 2.00 M
sulfuric acid reacts with excess potassium hydroxide.
Question 17
Question
A chemist wants to produce 200 grams of iron(III) oxide (F e2O3) through the
reaction of iron and oxygen gas. If the reaction is 80
Solution
Step 1: Write the balanced chemical equation for the reaction:
4F e + 3O2→2F e2O3
Step 2: Calculate the molar mass of F e2O3:
Molarmassof F e2O3= (2 ×M olarmassof F e) + (3 ×MolarmassofO)
14
= (2 ×55.85g/mol) + (3 ×16.00g/mol)
= 159.70g/mol
Step 3: Calculate the theoretical yield of F e2O3:
T heoreticalyield =200g
1×1mol
159.70g×4molF e
2molF e2O3
Step 4: Calculate the actual yield of F e2O3accounting for 80
Actualyield = 200g×0.80 = 160g
Step 5: Use the actual yield to calculate the grams of iron required:
160g=xg ×1molF e
55.85g
x= 90.25g
Therefore, the chemist needs 90.25 grams of iron to produce 200 grams of
iron(III) oxide.
Question 18
Question
A sample of magnesium reacts with sulfuric acid to produce hydrogen gas and
magnesium sulfate. If 3.00 grams of magnesium reacts with excess sulfuric acid
to produce 8.62 liters of hydrogen gas at STP, what is the percent yield of the
reaction?
(Atomic masses: Mg = 24.31 g/mol, S = 32.07 g/mol, O = 16.00 g/mol, H
= 1.01 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between magnesium and sulfuric acid is:
Mg + H2SO4→MgSO4+ H2
Step 2: Calculate the theoretical yield of hydrogen gas. First, calculate the
number of moles of magnesium:
moles of Mg = mass
molar mass =3.00
24.31 ≈0.123 moles
According to the balanced chemical equation, 1 mole of magnesium produces
1 mole of hydrogen gas. Therefore, the number of moles of hydrogen produced
will be the same as the number of moles of magnesium, which is 0.123 moles.
15
Next, calculate the volume of 0.123 moles of hydrogen gas at STP:
Volume = moles ×22.4 L/mol = 0.123 ×22.4≈2.75 L
So, the theoretical yield of hydrogen gas is 2.75 L.
Step 3: Calculate the percent yield of the reaction. Percent yield is calculated
using the formula:
Percent yield = Actual yield
Theoretical yield×100%
In this case, the actual yield is 8.62 L and the theoretical yield is 2.75 L.
Substituting these values into the formula:
Percent yield = 8.62
2.75×100% ≈313.09%
Therefore, the percent yield of the reaction is approximately 313.09
Question 19
Question
A mixture of aluminum and iodine weighs 15.0 g. When this mixture reacts,
it produces aluminum iodide. If the reaction goes to completion, what mass of
aluminum iodide is produced?
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
and iodine to form aluminum iodide. Step 2: Calculate the molar masses of alu-
minum, iodine, and aluminum iodide. Step 3: Determine the limiting reactant
using the given masses of aluminum and iodine. Step 4: Calculate the theoret-
ical yield of aluminum iodide based on the limiting reactant. Step 5: State the
mass of aluminum iodide produced.
Question 20
Question
A compound XCl2contains 32.4
Solution
Step 1: Calculate the molar mass of XCl2. Given that XCl2has a molar mass
of 200.0 g/mol, it can be expressed as:
MXCl2=MX+ 2MCl = 200.0 g/mol
16
where MXis the molar mass of element X and MCl is the molar mass of chlorine
(35.45 g/mol). Solving for MX, we get:
MX= 200.0−2(35.45) = 129.10 g/mol
Step 2: Determine the mass of chlorine in 1 mole of XCl2. Since the com-
pound XCl2contains 32.4
Mass of Cl = 0.324 ×200.0 g = 64.80 g
Step 3: Calculate the moles of chlorine in 1 mole of XCl2. Using the molar
mass of chlorine, we find the moles of chlorine in 1 mole of XCl2:
Moles of Cl = 64.80 g
35.45 g/mol ≈1.83 mol
Step 4: Determine the moles of element X in 1 mole of XCl2. From the
chemical formula of XCl2, we know that there is 1 mole of element X in 1 mole
of the compound. Therefore, the moles of element X in 1 mole of XCl2is also
1.
Step 5: Calculate the molar mass of element X. Since the molar mass of
element X is equal to its mass (129.10 g/mol), the identity of element X is the
element with a molar mass of 129.10 g/mol.
Question 21
Question
A student performs a synthesis reaction that produces 20.0 g of an unknown
compound. The balanced chemical equation for the reaction is:
2A+ 3B→C
If the student started with 15.0 g of compound A and 25.0 g of compound
B, determine the limiting reactant and the theoretical yield of compound C in
grams.
Solution
Step 1: Calculate the number of moles of each reactant.
Given: - Mass of compound A = 15.0 g - Molar mass of compound A = 50.0
g/mol - Mass of compound B = 25.0 g - Molar mass of compound B = 25.0
g/mol
Number of moles of compound A:
moles(A) = mass(A)
molar mass(A)=15.0 g
50.0 g/mol = 0.30 mol
17
Number of moles of compound B:
moles(B) = mass(B)
molar mass(B)=25.0 g
25.0 g/mol = 1.00 mol
Step 2: Determine the limiting reactant.
Compare the moles of each reactant to the stoichiometry of the reaction. The
reactant that produces the least amount of product is the limiting reactant.
From the balanced chemical equation, we see that 2 moles of A react with 3
moles of B to produce 1 mole of C.
Using compound A:
moles(A)
2=0.30 mol
2= 0.15 mol of C produced
Using compound B:
moles(B)
3=1.00 mol
3≈0.33 mol of C produced
Since compound A produces 0.15 mol of product while compound B produces
0.33 mol of product, compound A is the limiting reactant.
Step 3: Calculate the theoretical yield of compound C.
Using compound A as the limiting reactant:
moles of C produced = 0.15 mol
Molar mass of compound C is unknown, so we cannot directly convert moles
to grams at this point. We would need to know the molar mass of compound C
to find the theoretical yield in grams.
Question 22
Question
A reaction of 4.5 moles of hydrogen gas with excess nitrogen gas produces 13.5
moles of ammonia gas. Determine the balanced chemical equation for the reac-
tion and calculate the number of moles of nitrogen gas consumed in the reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction. Let’s assume
the balanced chemical equation for the reaction is:
aH2+bN2→cNH3
Given that 4.5 moles of hydrogen gas react with excess nitrogen gas to pro-
duce 13.5 moles of ammonia gas, we can set up a ratio of moles based on the
coefficients in the balanced chemical equation:
4.5
a=13.5
c
18
Since the stoichiometric coefficients must be in the simplest whole number ratio,
we can determine that:
a= 3, c = 9
Therefore, the balanced chemical equation is:
3H2+N2→2NH3
Step 2: Calculate the number of moles of nitrogen gas consumed in the
reaction. From the balanced chemical equation, we can see that 1 mole of
nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia
gas. Given that 4.5 moles of hydrogen gas were used, the number of moles of
nitrogen gas consumed can be calculated as follows:
Moles of N2=4.5 moles H2
3= 1.5 moles N2
Therefore, 1.5 moles of nitrogen gas were consumed in the reaction.
Question 23
Question
A student wanted to determine the amount of lead(II) chloride that could be
produced by reacting 25.0 g of lead(II) nitrate with excess sodium chloride
according to the balanced chemical equation:
Pb(NO3)2+ 2NaCl →PbCl2+ 2NaNO3
If the lead(II) chloride is recovered with a yield of 85.0
Solution
Step 1: Calculate the molar mass of lead(II) nitrate (Pb(NO3)2) and lead(II)
chloride (PbCl2). - The molar mass of Pb(NO3)2is calculated as follows:
Pb(NO3)2= Pb + 2(N + 3O) = 207.2 g/mol
- The molar mass of PbCl2is calculated as follows:
PbCl2= Pb + 2Cl = 278.1 g/mol
Step 2: Determine the number of moles of Pb(NO3)2used. - Using the given
mass of Pb(NO3)2:
Moles of Pb(NO3)2=25.0 g
207.2 g/mol = 0.121 mol
Step 3: Determine the theoretical yield of PbCl2using stoichiometry. -
According to the balanced chemical equation, 1 mol of Pb(NO3)2produces 1
mol of PbCl2. - Therefore, the moles of PbCl2produced is also 0.121 mol.
19
Step 4: Calculate the expected mass of PbCl2based on the theoretical yield
and percent yield.
Expected mass of PbCl2= 0.121 mol ×278.1 g/mol ×0.850 = 27.99 g
The student would expect to produce approximately 27.99 g of lead(II) chlo-
ride.
Question 24
Question
A compound containing only carbon, hydrogen, and nitrogen is analyzed and
found to consist of 60.00
Solution
Step 1: Find the molar mass of the empirical formula.
To find the empirical formula, we first need to find the mole ratios of the
elements in the compound. Assume we have 100 g of the compound. - The
number of moles of carbon = 60.00g
12.01g/mol = 4.996 mol - The number of moles
of hydrogen = 13.64g
1.008g/mol = 13.53 mol - The number of moles of nitrogen =
26.36g
14.01g/mol = 1.881 mol
Step 2: Find the mole ratios of the elements and determine the empirical
formula.
The mole ratio of carbon to hydrogen to nitrogen is approximately 5:13:2.
Step 3: Determine the empirical formula.
Since the molar ratio of carbon, hydrogen, and nitrogen is 5:13:2, the em-
pirical formula can be represented as C5H13N2.
Step 4: Find the empirical formula molar mass.
The molar mass of the empirical formula C5H13N2is 5(12.01g/mol)+13(1.008g/mol)+
2(14.01g/mol) = 87.16 g/mol.
Step 5: Determine the multiple required to obtain the molecular formula.
Divide the given molar mass of the compound by the molar mass of the
empirical formula to find the multiple required. 90g/mol
87.16g/mol ≈1.032.
Step 6: Determine the molecular formula.
Multiply the subscripts in the empirical formula by the multiple obtained:
1.032 ×C5H13N2≈C5.16H13.4N2.064.
Since we can’t have fractions in a molecular formula, we need to round to
the nearest whole number. Therefore, the molecular formula is C5H13N2.
20
Question 25
Question
Calculate the mass of iron(III) oxide that can be produced from the reaction of
75.0 g of iron with excess oxygen gas.
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between iron and oxygen to produce iron(III)
oxide is:
4 Fe + 3 O2→2 Fe2O3
Step 2: Calculate the molar mass of Fe and Fe2O3. The molar mass of Fe is
55.85 g/mol. The molar mass of Fe2O3is calculated as:
2×molar mass of Fe+3×molar mass of O = 2×55.85 g/mol+3×16.00 g/mol = 159.70 g/mol
Step 3: Determine the number of moles of iron present. Given mass of iron,
m= 75.0 g Molar mass of Fe, Molar massFe = 55.85 g/mol Number of moles of
Fe,
molesFe =m
Molar massFe
=75.0 g
55.85 g/mol = 1.342 mol
Step 4: Determine the limiting reactant. From the balanced chemical equa-
tion, the stoichiometry ratio of Fe to Fe2O3is 4:2 or simplified to 2:1. Therefore,
1.342 moles of Fe will produce 1.342
2= 0.671 moles of Fe2O3.
Step 5: Calculate the mass of iron(III) oxide produced. Number of moles of
Fe2O3= 0.671 mol Molar mass of Fe2O3= 159.70 g/mol (from Step 2) Mass
of Fe2O3produced,
mass of Fe2O3= moles of Fe2O3×Molar massFe2O3= 0.671 mol×159.70 g/mol = 107.04 g
Therefore, 107.04 g of iron(III) oxide can be produced from the reaction.
Question 26
Question
A compound X contains only C, H, and N. When a 1.50 g sample of X is burned
in excess oxygen, 4.56 g of CO2 and 1.48 g of H2O are produced. In another
experiment, when a 0.500 g sample of X is analyzed, it yields 0.312 g of NH3.
Determine the empirical formula of X.
21
Solution
Step 1: Calculate moles of carbon dioxide produced. Given the molar masses:
CO2 = 44.01 g/mol, H2O = 18.02 g/mol
Moles of CO2 = 4.56 g
44.01 g/mol = 0.1037 mol
Step 2: Calculate moles of water produced.
Moles of H2O = 1.48 g
18.02 g/mol = 0.0821 mol
Step 3: Find moles of carbon and hydrogen in the compound X. Let the
moles of C and H be n(C) and n(H) respectively.
From the combustion of X: 1 mol of C produces 1 mol of CO2, and 1 mol of
H produces 0.5 mol of H2O.
So, we have:
n(C) = 0.1037 mol
n(H) = 2 ×0.0821 mol = 0.1642 mol
Step 4: Calculate the moles of nitrogen in the compound X. From the anal-
ysis of X:
Moles of NH3 = 0.312 g
17.03 g/mol = 0.0183 mol
Step 5: Determine the empirical formula. The compound X has 0.1037 moles
of C, 0.1642 moles of H, and 0.0183 moles of N.
Dividing by the smallest number of moles (0.0183):
n(C) = 0.1037
0.0183 ≈5.67
n(H) = 0.1642
0.0183 ≈8.97
n(N) = 0.0183
0.0183 = 1
Step 6: Write the empirical formula. The empirical formula of the compound
X is C6H9N.
Question 27
Question
A reaction between iron (III) oxide and carbon monoxide produces iron metal
and carbon dioxide.
F e2O3(s)+3CO(g)→2F e(s)+3CO2(g)
If 256 grams of iron (III) oxide and 64 grams of carbon monoxide are available
for the reaction, determine: (a) The limiting reactant (b) The theoretical yield
of iron (c) The percentage yield if 115 grams of iron are actually produced.
22
Solution
Step 1: Calculate the moles of each reactant. Given: - Mass of iron (III) oxide
= 256 g - Molar mass of iron (III) oxide (F e2O3) = 159.69 g/mol - Mass of
carbon monoxide = 64 g - Molar mass of carbon monoxide (CO) = 28.01 g/mol
(a) The limiting reactant is the one that is completely consumed and limits
the amount of product formed.
Step 2: Calculate the moles of iron (III) oxide and carbon monoxide. For
iron (III) oxide:
Moles = Mass
Molar mass =256 g
159.69 g/mol ≈1.604 mol
For carbon monoxide:
Moles = Mass
Molar mass =64 g
28.01 g/mol ≈2.285 mol
Step 3: Determine the limiting reactant. From the balanced chemical equa-
tion, the stoichiometry ratio of F e2O3to CO is 1:3. Using the moles calculated
above, we can see that iron (III) oxide produces 1.604 mol×3 = 4.812 mol of car-
bon monoxide, which is more than the available 2.285 mol of carbon monoxide.
Therefore, iron (III) oxide is the limiting reactant.
(b) The theoretical yield of iron can be calculated using the limiting reactant.
Step 4: Calculate the theoretical yield of iron. From the balanced chemical
equation, the stoichiometry ratio of F e2O3to F e is 1:2. Theoretical yield of
iron:
Moles of F e = 1.604 mol ×2 mol F e
1 mol F e2O3
= 3.208 mol
Step 5: Convert moles of iron to grams.
Mass of F e = 3.208 mol ×55.85 g/mol = 179.67 g
(c) The percentage yield is calculated using the actual yield and theoretical
yield.
Step 6: Calculate the percentage yield. Given: - Actual yield of iron = 115
g
Percentage yield:
% Yield = Actual yield
Theoretical yield×100% = 115 g
179.67 g×100% ≈63.96%
Therefore, the percentage yield of iron is approximately 63.96
Question 28
Question
A sample of iron (III) oxide, Fe2O3, with a mass of 25.0 grams is reacted with
an excess of carbon monoxide, CO, in the following unbalanced equation:
Fe2O3+ CO →Fe + CO2
23
If the reaction produces 9.0 grams of iron, what is the theoretical yield of carbon
dioxide, CO2, in grams? (Molar masses: Fe2O3= 159.69 g/mol, Fe = 55.85
g/mol, CO = 28.01 g/mol, CO2= 44.01 g/mol)
Solution
Step 1: Find the moles of iron produced. Given: Mass of iron produced, m =
9.0 g; Molar mass of iron, M = 55.85 g/mol. We can use the formula n=m
Mto
find the number of moles of iron produced:
n=9.0 g
55.85 g/mol
n≈0.161 mol
Step 2: Calculate the moles of Fe2O3reacted. Using the balanced chemical
equation, we see that 1 mole of Fe2O3should produce 2 moles of Fe. Therefore,
the moles of Fe2O3reacted is half the moles of iron produced.
nFe2O3=1
2×0.161 mol
nFe2O3= 0.0805 mol
Step 3: Calculate the moles of CO2produced. From the balanced chemical
equation, 1 mole of Fe2O3reacts with 1 mole of CO to produce 1 mole of CO2.
Therefore, the moles of CO2produced would be the same as the moles of Fe2O3
reacted.
nCO2= 0.0805 mol
Step 4: Find the mass of CO2produced. The mass of CO2produced can be
calculated using the formula m=n×M:
m= 0.0805 mol ×44.01 g/mol
m≈3.54 g
Therefore, the theoretical yield of carbon dioxide, CO2, in this reaction is
approximately 3.54 grams.
Question 29
Question
Calculate the mass of ammonium nitrate that can be produced from the reaction
of 50.0 g of ammonia gas (NH3) and 75.0 g of nitric acid (HNO3) according to
the following balanced chemical equation:
NH3+HNO3→N H4N O3
24
Solution
Step 1: Write the balanced chemical equation for the reaction.
NH3+HNO3→N H4N O3
Step 2: Calculate the number of moles of each reactant. Given: - Mass
of ammonia gas (NH3) = 50.0 g - Molar mass of ammonia gas (NH3) = 17.03
g/mol - Mass of nitric acid (HNO3) = 75.0 g - Molar mass of nitric acid (HNO3)
= 63.01 g/mol
Number of moles of ammonia gas:
moles NH3=50.0 g
17.03 g/mol = 2.94 mol
Number of moles of nitric acid:
moles HNO3=75.0 g
63.01 g/mol = 1.19 mol
Step 3: Determine the limiting reactant. From the balanced chemical equa-
tion, we see that 1 mol of NH3reacts with 1 mol of HNO3to produce 1 mol of
NH4NO3. To determine the limiting reactant, we need to compare the ratios of
moles of reactants to stoichiometric coefficients. The ratio of moles of NH3to
HNO3is: 2.94 mol NH3
1.19 mol HNO3
≈2.47
So, HNO3is the limiting reactant.
Step 4: Calculate the theoretical yield of ammonium nitrate. From the
reaction equation, 1 mol of NH4NO3has a molar mass of:
14.01 + 4(1.008) + 14.01 + 3(16.00) = 80.04 g/mol
moles NH4NO3= 1.19 mol HNO3×1 mol NH4NO3
1 mol HNO3= 1.19 mol
The theoretical yield of NH4NO3is:
mass NH4NO3= 1.19 mol ×80.04 g/mol = 95.2 g
Therefore, the mass of ammonium nitrate that can be produced is 95.2 g.
Question 30
Question
A researcher is studying a reaction that produces nitrogen monoxide gas. The
reaction is represented by the following balanced chemical equation:
4NH3(g) + 5O2(g)→4NO(g) + 6H2O(g)
If 5.00 moles of ammonia (NH3) react with excess oxygen (O2), how many moles
of water (H2O) will be produced?
25
Question 3
Question
A reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) pro-
duces potassium sulfate (K2SO4) and water. If 50.0 mL of 0.500 M sulfuric
acid is reacted with excess potassium hydroxide, calculate the volume of 1.00
M potassium hydroxide needed to completely react with the sulfuric acid.
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid and potassium hydroxide.
H2SO4+ 2KOH →K2SO4+ 2H2O
Step 2: Determine the number of moles of sulfuric acid using the given
concentration and volume.
moles of H2SO4= Molarity ×Volume (L)
moles of H2SO4= 0.500 mol/L ×0.0500 L
moles of H2SO4= 0.0250 mol
Step 3: Use the mole ratio from the balanced equation to find the moles of
potassium hydroxide needed. From the equation, 1 mole of H2SO4reacts with
2 moles of KOH. Therefore, moles of KOH = 2 ×moles of H2SO4.
moles of KOH = 2 ×0.0250 mol
moles of KOH = 0.0500 mol
Step 4: Calculate the required volume of 1.00 M potassium hydroxide using
the moles and concentration.
Molarity = moles
volume (L) ⇒volume = moles
Molarity
volume of 1.00 M KOH = 0.0500 mol
1.00 mol/L
volume of 1.00 M KOH = 0.0500 L = 50.0 mL
Therefore, 50.0 mL of 1.00 M potassium hydroxide is needed to completely
react with 50.0 mL of 0.500 M sulfuric acid.
Question 4
Question
A sample of iron(III) chloride, FeCl3, is found to contain 2.50 moles of chlorine
atoms. How many moles of iron atoms are present in the sample?
3
Solution
Step 1: Write the balanced chemical equation for the dissociation of iron(III)
chloride.
FeCl3→Fe3+ + 3Cl−
Step 2: Determine the mole ratio between chlorine atoms and iron atoms
from the balanced chemical equation. For every mole of iron(III) chloride, there
are 1 mole of iron atoms and 3 moles of chlorine atoms. So, the mole ratio
between iron atoms and chlorine atoms is 1:3.
Step 3: Use the mole ratio to find the number of moles of iron atoms. Since
we have 2.50 moles of chlorine atoms, we can set up the following proportion:
2.50 mol Cl
3=xmol Fe
1
Step 4: Solve for the number of moles of iron atoms:
x=2.50 ×1
3= 0.833 mol Fe
Therefore, there are 0.833 moles of iron atoms present in the sample of
iron(III) chloride.
Question 5
Question
A student is performing a reaction in the lab that produces 2 moles of product
F for every 3 moles of reactant R. If the student starts with 4 moles of reactant
R, how many moles of product F will be produced?
Solution
Step 1: Determine the mole ratio between reactant R and product F. The mole
ratio between reactant R and product F is 3 moles of R to 2 moles of F.
Step 2: Calculate the moles of product F produced from 4 moles of reactant
R. Given that there are 4 moles of reactant R, we can set up a proportion to
find the moles of product F produced:
4 moles R
3=xmoles of F
2
Step 3: Solve for x to find the moles of product F. Cross multiplying, we
get:
4×2=3x
8=3x
x=8
3=2
3×8 = 16
3
Thus, 16/3 moles of product F will be produced when 4 moles of reactant R are
used.
4
Question 6
Question
A chemist wishes to produce 250.0 g of iron (III) oxide according to the following
balanced chemical equation:
4F e + 3O2→2F e2O3
If iron is available as iron (III) chloride, FeCl3, how many grams of iron (III)
chloride should be used to produce the desired amount of iron (III) oxide?
Solution
Step 1: Find the molar mass of iron (III) oxide (F e2O3). The molar mass of
F e2O3is calculated by adding the molar mass of iron (Fe) and three times the
molar mass of oxygen (O).
= 2 ×Atomic mass of Fe + 3 ×Atomic mass of O
= 2 ×55.85 g/mol + 3 ×16.00 g/mol
= 159.70 g/mol
Step 2: Calculate the number of moles of iron (III) oxide needed. Given
mass of iron (III) oxide, mF e2O3= 250.0 g. Using the molar mass calculated
above, we can find the number of moles of F e2O3using the formula:
n=m
M
n=250.0 g
159.70 g/mol
n≈1.567 mol
Step 3: Determine the mole ratio between iron (III) chloride and iron (III)
oxide. From the balanced chemical equation, the mole ratio between iron (III)
chloride (FeCl3) and iron (III) oxide (F e2O3) is 4:2. This can be simplified to
a ratio of 2:1.
Step 4: Convert moles of iron (III) oxide to moles of iron (III) chloride. Since
the mole ratio is 2:1, the number of moles of iron (III) chloride needed is half
the number of moles of iron (III) oxide:
nF eCl3=1.567 mol
2
nF eCl3= 0.784 mol
Step 5: Calculate the mass of iron (III) chloride required. Using the molar
mass of iron (III) chloride (F eCl3):
MF eCl3= 162.20 g/mol
5
The mass of iron (III) chloride needed can be calculated using the formula:
mF eCl3=nF eCl3×MF eCl3
mF eCl3= 0.784 mol ×162.20 g/mol
mF eCl3≈127.24 g
Therefore, approximately 127.24 grams of iron (III) chloride should be used
to produce 250.0 grams of iron (III) oxide.
Question 7
Question
A chemist is studying the reaction between silver nitrate (AgNO3) and sodium
chloride (NaCl). If 10.0 grams of silver nitrate reacts with excess sodium chlo-
ride, what is the theoretical yield of silver chloride that can be produced? The
balanced chemical equation for the reaction is:
AgNO3+ NaCl →AgCl + NaNO3
Solution
Step 1: Determine the molar mass of silver nitrate and calculate the number of
moles. The molar mass of AgNO3is:
1Ag + 1N + 3O = 107.87g/mol + 14.01g/mol + 3(16.00g/mol) = 169.87g/mol
The number of moles of silver nitrate is calculated as:
moles = mass
molar mass =10.0g
169.87g/mol = 0.0589 mol
Step 2: Use the balanced chemical equation to determine the stoichiometry
of the reaction. From the balanced chemical equation, we see that 1 mole of
AgNO3produces 1 mole of AgCl.
Step 3: Calculate the theoretical yield of silver chloride. Since the stoichio-
metric ratio is 1:1, the number of moles of AgCl produced will be the same as
the number of moles of AgNO3used.
The mass of silver chloride produced is calculated as:
mass = moles ×molar mass = 0.0589 mol ×143.32 g/mol = 8.46 g
Therefore, the theoretical yield of silver chloride that can be produced is
8.46 grams.
6
Question 8
Question
A student wants to determine the amount of potassium permanganate, KM nO4,
needed to titrate a 0.500 M solution of iron(II) ion, F e2+, in acidic solution.
The balanced chemical equation for the reaction is:
5F e2+ + 8H++MnO−
4→5F e3+ +Mn2+ + 4H2O
If it takes 40.0 mL of the potassium permanganate solution to reach the equiv-
alence point, what mass of KMnO4in grams would be needed?
(Hint: The molar mass of KM nO4is 158.03 g/mol)
Solution
Step 1: Determine the moles of F e2+ in the solution. Given that the volume of
the F e2+ solution is 40.0 mL and its concentration is 0.500 M, we can calculate
the moles of F e2+.
Moles of F e2+ = Volume×Concentration = 40.0×10−3L×0.500 mol/L = 0.0200 mol
Step 2: Use the stoichiometry of the balanced equation to determine the
moles of KMnO4needed. From the balanced chemical equation, we see that 1
mole of MnO−
4reacts with 5 moles of F e2+.
Moles of KMnO4=Moles of F e2+
5=0.0200 mol
5= 0.00400 mol
Step 3: Calculate the mass of KM nO4needed. Using the molar mass of
KMnO4(158.03 g/mol), we can find the mass of KM nO4required.
Mass of KMnO4= Moles×Molar Mass = 0.00400 mol×158.03 g/mol = 0.632 g
Therefore, the mass of KMnO4needed to titrate the F e2+ solution is 0.632
grams.
Question 9
Question
A sample of an unknown compound containing carbon, hydrogen, and oxygen
is analyzed. Combustion of a 0.568 g sample of the compound yields 1.105 g
of carbon dioxide and 0.471 g of water. What is the empirical formula of the
compound?
(Molar masses: C = 12.011 g/mol, H = 1.008 g/mol, O = 15.999 g/mol)
7
Solution
Step 1: Find the moles of carbon and hydrogen in the compound. - The molar
mass of carbon dioxide is 44.01 g/mol, so the moles of carbon dioxide produced
can be calculated as:
moles of CO2=1.105 g
44.01 g/mol = 0.0251 mol
- Since each mole of carbon dioxide contains 1 mole of carbon, the moles of
carbon in the compound is also 0.0251 mol.
- The molar mass of water is 18.015 g/mol, so the moles of water produced
can be calculated as:
moles of H2O=0.471 g
18.015 g/mol = 0.0261 mol
- Since each mole of water contains 2 moles of hydrogen, the moles of hydro-
gen in the compound is 2 * 0.0261 mol = 0.0522 mol.
Step 2: Determine the molar ratios of carbon, hydrogen, and oxygen in the
compound. - The ratio of carbon to hydrogen is 0.0251 mol : 0.0522 mol, which
simplifies to 1 : 2.08.
Step 3: Convert the molar ratios to whole numbers. - The molar ratio of
carbon to hydrogen can be approximated to 1 : 2.
Step 4: Write the empirical formula using the whole-number ratios. - The
empirical formula of the compound is CH2.
Question 10
Question
A sample of aluminum chloride (AlCl3) reacts with excess sodium carbonate
(Na2CO3) to produce aluminum carbonate (Al2(CO3)3) and sodium chloride
(NaCl). If 15.0 grams of aluminum chloride are reacted with 25.0 grams of
sodium carbonate, what mass of aluminum carbonate will be produced?
(Molar masses: AlCl3= 133.34 g/mol, Na2CO3= 105.99 g/mol, Al2(CO3)3=
324.95 g/mol, NaCl = 58.44 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
chloride and sodium carbonate.
2AlCl3+ 3N a2CO3→Al2(CO3)3+ 6N aCl
8
Step 2: Calculate the number of moles of each reactant.
Moles of AlCl3=15.0 g
133.34 g/mol
= 0.1124 mol
Moles of Na2CO3=25.0 g
105.99 g/mol
= 0.2360 mol
Step 3: Determine the limiting reactant by examining the mole ratio between
aluminum chloride and sodium carbonate from the balanced equation.
Moles of AlCl3
2=0.1124 mol
2= 0.0562 mol
Moles of Na2CO3
3=0.2360 mol
3= 0.0787 mol
Since aluminum chloride produces the least amount of product, it is the limiting
reactant.
Step 4: Calculate the theoretical yield of aluminum carbonate using the
limiting reactant.
Moles of Al2(CO3)3= 0.0562 mol ×1 mol Al2(CO3)3
2 mol AlCl3
= 0.0281 mol
Mass of Al2(CO3)3= 0.0281 mol ×324.95 g/mol
= 9.13 g
Answer: The mass of aluminum carbonate produced is 9.13 grams.
Question 11
Question
A reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) pro-
duces sodium sulfate (Na2SO4) and water (H2O).
Given a 1.50 L solution of H2SO4with a concentration of 0.500 M, determine
the volume of 2.00 M NaOH solution needed to completely react with the
sulfuric acid. Assume that the reaction goes to completion and that the volumes
of the solutions are additive.
Solution
Step 1: Write the balanced chemical equation for the reaction between H2SO4
and NaOH.
H2SO4(aq)+2NaOH(aq)→N a2SO4(aq)+2H2O(l)
9
Step 2: Determine the number of moles of H2SO4present in the solution.
Given: Volume of H2SO4solution = 1.50 L Concentration of H2SO4solution
= 0.500 M
Using the formula C=n
V(where C= concentration, n= moles, and V=
volume), we have: n=C×V n = 0.500 mol/L ×1.50 L n= 0.750 moles
Step 3: Use the stoichiometry of the balanced chemical equation to find the
amount of NaOH needed to react with all the H2SO4. From the balanced
chemical equation, we see that 1 mole of H2SO4reacts with 2 moles of NaOH.
Therefore, to react completely with 0.750 moles of H2SO4, we need: 2 ×
0.750 = 1.50 moles of NaOH
Step 4: Calculate the volume of 2.00 M NaOH solution needed. Concen-
tration of NaOH solution = 2.00 M Volume of NaOH solution needed can be
calculated using the formula: V=n
C(where V= volume, n= moles, and C=
concentration)
We have: V=1.50 moles
2.00 mol/L V= 0.75 L
Therefore, 0.75 L of 2.00 M NaOH solution is needed to completely react
with the sulfuric acid.
Question 12
Question
A sample of calcium carbonate, CaCO3, with a mass of 5.00 g, reacts completely
with excess hydrochloric acid solution according to the following balanced chem-
ical equation:
CaCO3(s) + 2HCl(aq)→CaCl2(aq) + CO2(g)+H2O(l)
Calculate the volume of carbon dioxide produced at STP.
Solution
Step 1: Calculate the number of moles of calcium carbonate.
Molar mass of CaCO3= 40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol)
= 100.09 g/mol
Number of moles of CaCO3=Mass
Molar mass
=5.00 g
100.09 g/mol
= 0.04997 mol
Step 2: Use the balanced chemical equation to relate the moles of carbon
dioxide produced to the moles of calcium carbonate. According to the balanced
10
chemical equation, 1 mole of CaCO3produces 1 mole of CO2. Thus, 0.04997
mol of CaCO3will produce 0.04997 mol of CO2.
Step 3: Calculate the volume of carbon dioxide produced at STP. At STP
(Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.
Volume of CO2= Number of moles ×22.4 L/mol
= 0.04997 mol ×22.4 L/mol
= 1.12 L
Therefore, the volume of carbon dioxide produced at STP is 1.12 liters.
Question 13
Question
A chemist is trying to synthesize a compound using the following reaction:
2C6H6+ 15O2→12CO2+ 6H2O
If the chemist has 200 grams of benzene (C6H6) and plenty of oxygen gas (O2),
how many grams of water (H2O) will be produced in theory?
Solution
Step 1: Find the molar mass of benzene (C6H6). The molar mass of carbon
(C) is 12.01 g/mol and the molar mass of hydrogen (H) is 1.008 g/mol. Since
benzene has 6 carbons and 6 hydrogens, the molar mass of benzene is:
6(12.01 g/mol) + 6(1.008 g/mol) = 78.11 g/mol
Step 2: Calculate the number of moles of benzene. Given that the chemist
has 200 grams of benzene:
200 g
78.11 g/mol = 2.563 mol
Step 3: Determine the limiting reagent between benzene and oxygen. From
the balanced chemical equation, we can see that 2 moles of benzene react with
15 moles of oxygen gas. Let’s calculate the number of moles of oxygen gas
needed to react with the given amount of benzene:
2.563 mol C6H6×15 mol O2
2 mol C6H6
= 19.218 mol O2
Step 4: Calculate the mass of water produced. From the balanced chemical
equation, we can see that 2 moles of water are produced for every 15 moles of
oxygen gas. Let’s calculate the mass of water produced:
19.218 mol O2×6 mol H2O
15 mol O2
×18.015 g/mol H2O = 137.09 g H2O
Therefore, in theory, 137.09 grams of water will be produced.
11
Question 14
Question
A compound containing only carbon, hydrogen, and oxygen is burned in oxygen
gas. In one experiment, a 0.500 g sample of the compound produced 0.964 g
of carbon dioxide and 0.396 g of water. Determine the empirical formula of the
compound.
Solution
Step 1: Calculate moles of carbon dioxide produced.
Moles of CO2=Mass of CO2
Molar mass of CO2
=0.964 g
44.01 g/mol = 0.0219 mol
Step 2: Calculate moles of water produced.
Moles of H2O=Mass of H2O
Molar mass of H2O=0.396 g
18.02 g/mol = 0.0220 mol
Step 3: Determine the moles of carbon in the original compound. Since all
the carbon in the compound ends up in the form of CO2, the moles of carbon
in the original compound = moles of CO2produced. Moles of carbon = 0.0219
mol
Step 4: Determine the moles of hydrogen in the original compound. Since
all the hydrogen in the compound ends up in the form of H2O, the moles
of hydrogen in the original compound = moles of H2Oproduced. Moles of
hydrogen = 0.0220 mol
Step 5: Calculate the moles of oxygen in the original compound. The moles
of oxygen in the original compound can be calculated using the fact that the
sum of moles of carbon, hydrogen, and oxygen in the compound is equal to the
total moles of the compound. Total moles of compound = moles of carbon +
moles of hydrogen + moles of oxygen Total moles of compound = 0.500 g /
molar mass of compound
Step 6: Determine the empirical formula of the compound. Now that we have
the moles of carbon, hydrogen, and oxygen in the compound, we can determine
the empirical formula. Divide the moles of each element by the smallest number
of moles obtained:
Carbon: 0.0219 mol/0.0219 mol = 1
Hydrogen: 0.0220 mol/0.0219 mol ≈1
Oxygen: (Total moles of compound −moles of carbon −moles of hydrogen)/0.0219 mol
Thus, the empirical formula of the compound is CH1Ox.
12
Question 15
Question
A reaction between silver nitrate (AgNO3) and sodium phosphate (Na3PO4)
produces silver phosphate precipitate (Ag3PO4) and sodium nitrate (NaNO3).
If 5.00 g of silver nitrate is reacted with an excess of sodium phosphate and 8.00
g of silver phosphate is isolated, what is the percent yield of the reaction?
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction is:
3AgNO3+ Na3PO4→Ag3PO4+ 3NaNO3
Step 2: Calculate the molar mass of silver nitrate (AgNO3).
Molar mass of AgNO3= molar massAg + molar massN+ 3 ×molar massO
= 107.87 g/mol + 14.01 g/mol + 3 ×16.00 g/mol = 169.87 g/mol
Step 3: Calculate the number of moles of silver nitrate used.
Moles of AgNO3=mass of AgNO3
molar mass of AgNO3
=5.00 g
169.87 g/mol = 0.0294 mol
Step 4: Determine the theoretical yield of silver phosphate. From the bal-
anced chemical equation, it is clear that the molar ratio of AgNO3to Ag3PO4
is 3:1. Therefore, the theoretical yield of Ag3PO4is:
Theoretical yield of Ag3PO4= 3×moles of AgNO3= 3×0.0294 mol = 0.0882 mol
Step 5: Calculate the molar mass of silver phosphate (Ag3PO4).
Molar mass of Ag3PO4= 3 ×molar massAg + molar massP+ 4 ×molar massO
= 3 ×107.87 g/mol + 30.97 g/mol + 4 ×16.00 g/mol = 418.62 g/mol
Step 6: Calculate the actual yield of silver phosphate.
Actual yield of Ag3PO4= 8.00 g
Step 7: Calculate the percent yield of the reaction.
Percent yield = Actual yield
Theoretical yield ×100% = 8.00 g
0.0882 mol ×418.62 g/mol ×100%
= 44.6%
Therefore, the percent yield of the reaction is 44.6
13
Question 16
Question
A chemical reaction between sulfuric acid (H2SO4) and potassium hydroxide
(KOH) produces potassium sulfate and water. If 25.0 mL of 2.00 M sulfuric
acid reacts with excess potassium hydroxide, what mass of potassium sulfate is
produced?
Solution
Step 1: Write the balanced chemical equation for the reaction.
H2SO4+ 2KOH →K2SO4+ 2H2O
Step 2: Determine the number of moles of sulfuric acid (H2SO4) used.
moles of H2SO4= volume ×molarity = 0.025 L ×2.00 mol/L = 0.050 mol
Step 3: Use the mole ratio from the balanced equation to find the number
of moles of potassium sulfate (K2SO4) produced.
moles of K2SO4=0.050 mol H2SO4×1 mol K2SO4
1 mol H2SO4
= 0.050 mol K2SO4
Step 4: Calculate the mass of potassium sulfate produced using the molar
mass of K2SO4.
m= moles ×molar mass = 0.050 mol ×174.259 g/mol = 8.71 g
Therefore, 8.71 g of potassium sulfate is produced when 25.0 mL of 2.00 M
sulfuric acid reacts with excess potassium hydroxide.
Question 17
Question
A chemist wants to produce 200 grams of iron(III) oxide (F e2O3) through the
reaction of iron and oxygen gas. If the reaction is 80
Solution
Step 1: Write the balanced chemical equation for the reaction:
4F e + 3O2→2F e2O3
Step 2: Calculate the molar mass of F e2O3:
Molarmassof F e2O3= (2 ×M olarmassof F e) + (3 ×MolarmassofO)
14
= (2 ×55.85g/mol) + (3 ×16.00g/mol)
= 159.70g/mol
Step 3: Calculate the theoretical yield of F e2O3:
T heoreticalyield =200g
1×1mol
159.70g×4molF e
2molF e2O3
Step 4: Calculate the actual yield of F e2O3accounting for 80
Actualyield = 200g×0.80 = 160g
Step 5: Use the actual yield to calculate the grams of iron required:
160g=xg ×1molF e
55.85g
x= 90.25g
Therefore, the chemist needs 90.25 grams of iron to produce 200 grams of
iron(III) oxide.
Question 18
Question
A sample of magnesium reacts with sulfuric acid to produce hydrogen gas and
magnesium sulfate. If 3.00 grams of magnesium reacts with excess sulfuric acid
to produce 8.62 liters of hydrogen gas at STP, what is the percent yield of the
reaction?
(Atomic masses: Mg = 24.31 g/mol, S = 32.07 g/mol, O = 16.00 g/mol, H
= 1.01 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between magnesium and sulfuric acid is:
Mg + H2SO4→MgSO4+ H2
Step 2: Calculate the theoretical yield of hydrogen gas. First, calculate the
number of moles of magnesium:
moles of Mg = mass
molar mass =3.00
24.31 ≈0.123 moles
According to the balanced chemical equation, 1 mole of magnesium produces
1 mole of hydrogen gas. Therefore, the number of moles of hydrogen produced
will be the same as the number of moles of magnesium, which is 0.123 moles.
15
Next, calculate the volume of 0.123 moles of hydrogen gas at STP:
Volume = moles ×22.4 L/mol = 0.123 ×22.4≈2.75 L
So, the theoretical yield of hydrogen gas is 2.75 L.
Step 3: Calculate the percent yield of the reaction. Percent yield is calculated
using the formula:
Percent yield = Actual yield
Theoretical yield×100%
In this case, the actual yield is 8.62 L and the theoretical yield is 2.75 L.
Substituting these values into the formula:
Percent yield = 8.62
2.75×100% ≈313.09%
Therefore, the percent yield of the reaction is approximately 313.09
Question 19
Question
A mixture of aluminum and iodine weighs 15.0 g. When this mixture reacts,
it produces aluminum iodide. If the reaction goes to completion, what mass of
aluminum iodide is produced?
Solution
Step 1: Write the balanced chemical equation for the reaction between aluminum
and iodine to form aluminum iodide. Step 2: Calculate the molar masses of alu-
minum, iodine, and aluminum iodide. Step 3: Determine the limiting reactant
using the given masses of aluminum and iodine. Step 4: Calculate the theoret-
ical yield of aluminum iodide based on the limiting reactant. Step 5: State the
mass of aluminum iodide produced.
Question 20
Question
A compound XCl2contains 32.4
Solution
Step 1: Calculate the molar mass of XCl2. Given that XCl2has a molar mass
of 200.0 g/mol, it can be expressed as:
MXCl2=MX+ 2MCl = 200.0 g/mol
16
where MXis the molar mass of element X and MCl is the molar mass of chlorine
(35.45 g/mol). Solving for MX, we get:
MX= 200.0−2(35.45) = 129.10 g/mol
Step 2: Determine the mass of chlorine in 1 mole of XCl2. Since the com-
pound XCl2contains 32.4
Mass of Cl = 0.324 ×200.0 g = 64.80 g
Step 3: Calculate the moles of chlorine in 1 mole of XCl2. Using the molar
mass of chlorine, we find the moles of chlorine in 1 mole of XCl2:
Moles of Cl = 64.80 g
35.45 g/mol ≈1.83 mol
Step 4: Determine the moles of element X in 1 mole of XCl2. From the
chemical formula of XCl2, we know that there is 1 mole of element X in 1 mole
of the compound. Therefore, the moles of element X in 1 mole of XCl2is also
1.
Step 5: Calculate the molar mass of element X. Since the molar mass of
element X is equal to its mass (129.10 g/mol), the identity of element X is the
element with a molar mass of 129.10 g/mol.
Question 21
Question
A student performs a synthesis reaction that produces 20.0 g of an unknown
compound. The balanced chemical equation for the reaction is:
2A+ 3B→C
If the student started with 15.0 g of compound A and 25.0 g of compound
B, determine the limiting reactant and the theoretical yield of compound C in
grams.
Solution
Step 1: Calculate the number of moles of each reactant.
Given: - Mass of compound A = 15.0 g - Molar mass of compound A = 50.0
g/mol - Mass of compound B = 25.0 g - Molar mass of compound B = 25.0
g/mol
Number of moles of compound A:
moles(A) = mass(A)
molar mass(A)=15.0 g
50.0 g/mol = 0.30 mol
17
Number of moles of compound B:
moles(B) = mass(B)
molar mass(B)=25.0 g
25.0 g/mol = 1.00 mol
Step 2: Determine the limiting reactant.
Compare the moles of each reactant to the stoichiometry of the reaction. The
reactant that produces the least amount of product is the limiting reactant.
From the balanced chemical equation, we see that 2 moles of A react with 3
moles of B to produce 1 mole of C.
Using compound A:
moles(A)
2=0.30 mol
2= 0.15 mol of C produced
Using compound B:
moles(B)
3=1.00 mol
3≈0.33 mol of C produced
Since compound A produces 0.15 mol of product while compound B produces
0.33 mol of product, compound A is the limiting reactant.
Step 3: Calculate the theoretical yield of compound C.
Using compound A as the limiting reactant:
moles of C produced = 0.15 mol
Molar mass of compound C is unknown, so we cannot directly convert moles
to grams at this point. We would need to know the molar mass of compound C
to find the theoretical yield in grams.
Question 22
Question
A reaction of 4.5 moles of hydrogen gas with excess nitrogen gas produces 13.5
moles of ammonia gas. Determine the balanced chemical equation for the reac-
tion and calculate the number of moles of nitrogen gas consumed in the reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction. Let’s assume
the balanced chemical equation for the reaction is:
aH2+bN2→cNH3
Given that 4.5 moles of hydrogen gas react with excess nitrogen gas to pro-
duce 13.5 moles of ammonia gas, we can set up a ratio of moles based on the
coefficients in the balanced chemical equation:
4.5
a=13.5
c
18
Since the stoichiometric coefficients must be in the simplest whole number ratio,
we can determine that:
a= 3, c = 9
Therefore, the balanced chemical equation is:
3H2+N2→2NH3
Step 2: Calculate the number of moles of nitrogen gas consumed in the
reaction. From the balanced chemical equation, we can see that 1 mole of
nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia
gas. Given that 4.5 moles of hydrogen gas were used, the number of moles of
nitrogen gas consumed can be calculated as follows:
Moles of N2=4.5 moles H2
3= 1.5 moles N2
Therefore, 1.5 moles of nitrogen gas were consumed in the reaction.
Question 23
Question
A student wanted to determine the amount of lead(II) chloride that could be
produced by reacting 25.0 g of lead(II) nitrate with excess sodium chloride
according to the balanced chemical equation:
Pb(NO3)2+ 2NaCl →PbCl2+ 2NaNO3
If the lead(II) chloride is recovered with a yield of 85.0
Solution
Step 1: Calculate the molar mass of lead(II) nitrate (Pb(NO3)2) and lead(II)
chloride (PbCl2). - The molar mass of Pb(NO3)2is calculated as follows:
Pb(NO3)2= Pb + 2(N + 3O) = 207.2 g/mol
- The molar mass of PbCl2is calculated as follows:
PbCl2= Pb + 2Cl = 278.1 g/mol
Step 2: Determine the number of moles of Pb(NO3)2used. - Using the given
mass of Pb(NO3)2:
Moles of Pb(NO3)2=25.0 g
207.2 g/mol = 0.121 mol
Step 3: Determine the theoretical yield of PbCl2using stoichiometry. -
According to the balanced chemical equation, 1 mol of Pb(NO3)2produces 1
mol of PbCl2. - Therefore, the moles of PbCl2produced is also 0.121 mol.
19
Step 4: Calculate the expected mass of PbCl2based on the theoretical yield
and percent yield.
Expected mass of PbCl2= 0.121 mol ×278.1 g/mol ×0.850 = 27.99 g
The student would expect to produce approximately 27.99 g of lead(II) chlo-
ride.
Question 24
Question
A compound containing only carbon, hydrogen, and nitrogen is analyzed and
found to consist of 60.00
Solution
Step 1: Find the molar mass of the empirical formula.
To find the empirical formula, we first need to find the mole ratios of the
elements in the compound. Assume we have 100 g of the compound. - The
number of moles of carbon = 60.00g
12.01g/mol = 4.996 mol - The number of moles
of hydrogen = 13.64g
1.008g/mol = 13.53 mol - The number of moles of nitrogen =
26.36g
14.01g/mol = 1.881 mol
Step 2: Find the mole ratios of the elements and determine the empirical
formula.
The mole ratio of carbon to hydrogen to nitrogen is approximately 5:13:2.
Step 3: Determine the empirical formula.
Since the molar ratio of carbon, hydrogen, and nitrogen is 5:13:2, the em-
pirical formula can be represented as C5H13N2.
Step 4: Find the empirical formula molar mass.
The molar mass of the empirical formula C5H13N2is 5(12.01g/mol)+13(1.008g/mol)+
2(14.01g/mol) = 87.16 g/mol.
Step 5: Determine the multiple required to obtain the molecular formula.
Divide the given molar mass of the compound by the molar mass of the
empirical formula to find the multiple required. 90g/mol
87.16g/mol ≈1.032.
Step 6: Determine the molecular formula.
Multiply the subscripts in the empirical formula by the multiple obtained:
1.032 ×C5H13N2≈C5.16H13.4N2.064.
Since we can’t have fractions in a molecular formula, we need to round to
the nearest whole number. Therefore, the molecular formula is C5H13N2.
20
Question 25
Question
Calculate the mass of iron(III) oxide that can be produced from the reaction of
75.0 g of iron with excess oxygen gas.
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation for the reaction between iron and oxygen to produce iron(III)
oxide is:
4 Fe + 3 O2→2 Fe2O3
Step 2: Calculate the molar mass of Fe and Fe2O3. The molar mass of Fe is
55.85 g/mol. The molar mass of Fe2O3is calculated as:
2×molar mass of Fe+3×molar mass of O = 2×55.85 g/mol+3×16.00 g/mol = 159.70 g/mol
Step 3: Determine the number of moles of iron present. Given mass of iron,
m= 75.0 g Molar mass of Fe, Molar massFe = 55.85 g/mol Number of moles of
Fe,
molesFe =m
Molar massFe
=75.0 g
55.85 g/mol = 1.342 mol
Step 4: Determine the limiting reactant. From the balanced chemical equa-
tion, the stoichiometry ratio of Fe to Fe2O3is 4:2 or simplified to 2:1. Therefore,
1.342 moles of Fe will produce 1.342
2= 0.671 moles of Fe2O3.
Step 5: Calculate the mass of iron(III) oxide produced. Number of moles of
Fe2O3= 0.671 mol Molar mass of Fe2O3= 159.70 g/mol (from Step 2) Mass
of Fe2O3produced,
mass of Fe2O3= moles of Fe2O3×Molar massFe2O3= 0.671 mol×159.70 g/mol = 107.04 g
Therefore, 107.04 g of iron(III) oxide can be produced from the reaction.
Question 26
Question
A compound X contains only C, H, and N. When a 1.50 g sample of X is burned
in excess oxygen, 4.56 g of CO2 and 1.48 g of H2O are produced. In another
experiment, when a 0.500 g sample of X is analyzed, it yields 0.312 g of NH3.
Determine the empirical formula of X.
21
Solution
Step 1: Calculate moles of carbon dioxide produced. Given the molar masses:
CO2 = 44.01 g/mol, H2O = 18.02 g/mol
Moles of CO2 = 4.56 g
44.01 g/mol = 0.1037 mol
Step 2: Calculate moles of water produced.
Moles of H2O = 1.48 g
18.02 g/mol = 0.0821 mol
Step 3: Find moles of carbon and hydrogen in the compound X. Let the
moles of C and H be n(C) and n(H) respectively.
From the combustion of X: 1 mol of C produces 1 mol of CO2, and 1 mol of
H produces 0.5 mol of H2O.
So, we have:
n(C) = 0.1037 mol
n(H) = 2 ×0.0821 mol = 0.1642 mol
Step 4: Calculate the moles of nitrogen in the compound X. From the anal-
ysis of X:
Moles of NH3 = 0.312 g
17.03 g/mol = 0.0183 mol
Step 5: Determine the empirical formula. The compound X has 0.1037 moles
of C, 0.1642 moles of H, and 0.0183 moles of N.
Dividing by the smallest number of moles (0.0183):
n(C) = 0.1037
0.0183 ≈5.67
n(H) = 0.1642
0.0183 ≈8.97
n(N) = 0.0183
0.0183 = 1
Step 6: Write the empirical formula. The empirical formula of the compound
X is C6H9N.
Question 27
Question
A reaction between iron (III) oxide and carbon monoxide produces iron metal
and carbon dioxide.
F e2O3(s)+3CO(g)→2F e(s)+3CO2(g)
If 256 grams of iron (III) oxide and 64 grams of carbon monoxide are available
for the reaction, determine: (a) The limiting reactant (b) The theoretical yield
of iron (c) The percentage yield if 115 grams of iron are actually produced.
22
Solution
Step 1: Calculate the moles of each reactant. Given: - Mass of iron (III) oxide
= 256 g - Molar mass of iron (III) oxide (F e2O3) = 159.69 g/mol - Mass of
carbon monoxide = 64 g - Molar mass of carbon monoxide (CO) = 28.01 g/mol
(a) The limiting reactant is the one that is completely consumed and limits
the amount of product formed.
Step 2: Calculate the moles of iron (III) oxide and carbon monoxide. For
iron (III) oxide:
Moles = Mass
Molar mass =256 g
159.69 g/mol ≈1.604 mol
For carbon monoxide:
Moles = Mass
Molar mass =64 g
28.01 g/mol ≈2.285 mol
Step 3: Determine the limiting reactant. From the balanced chemical equa-
tion, the stoichiometry ratio of F e2O3to CO is 1:3. Using the moles calculated
above, we can see that iron (III) oxide produces 1.604 mol×3 = 4.812 mol of car-
bon monoxide, which is more than the available 2.285 mol of carbon monoxide.
Therefore, iron (III) oxide is the limiting reactant.
(b) The theoretical yield of iron can be calculated using the limiting reactant.
Step 4: Calculate the theoretical yield of iron. From the balanced chemical
equation, the stoichiometry ratio of F e2O3to F e is 1:2. Theoretical yield of
iron:
Moles of F e = 1.604 mol ×2 mol F e
1 mol F e2O3
= 3.208 mol
Step 5: Convert moles of iron to grams.
Mass of F e = 3.208 mol ×55.85 g/mol = 179.67 g
(c) The percentage yield is calculated using the actual yield and theoretical
yield.
Step 6: Calculate the percentage yield. Given: - Actual yield of iron = 115
g
Percentage yield:
% Yield = Actual yield
Theoretical yield×100% = 115 g
179.67 g×100% ≈63.96%
Therefore, the percentage yield of iron is approximately 63.96
Question 28
Question
A sample of iron (III) oxide, Fe2O3, with a mass of 25.0 grams is reacted with
an excess of carbon monoxide, CO, in the following unbalanced equation:
Fe2O3+ CO →Fe + CO2
23
If the reaction produces 9.0 grams of iron, what is the theoretical yield of carbon
dioxide, CO2, in grams? (Molar masses: Fe2O3= 159.69 g/mol, Fe = 55.85
g/mol, CO = 28.01 g/mol, CO2= 44.01 g/mol)
Solution
Step 1: Find the moles of iron produced. Given: Mass of iron produced, m =
9.0 g; Molar mass of iron, M = 55.85 g/mol. We can use the formula n=m
Mto
find the number of moles of iron produced:
n=9.0 g
55.85 g/mol
n≈0.161 mol
Step 2: Calculate the moles of Fe2O3reacted. Using the balanced chemical
equation, we see that 1 mole of Fe2O3should produce 2 moles of Fe. Therefore,
the moles of Fe2O3reacted is half the moles of iron produced.
nFe2O3=1
2×0.161 mol
nFe2O3= 0.0805 mol
Step 3: Calculate the moles of CO2produced. From the balanced chemical
equation, 1 mole of Fe2O3reacts with 1 mole of CO to produce 1 mole of CO2.
Therefore, the moles of CO2produced would be the same as the moles of Fe2O3
reacted.
nCO2= 0.0805 mol
Step 4: Find the mass of CO2produced. The mass of CO2produced can be
calculated using the formula m=n×M:
m= 0.0805 mol ×44.01 g/mol
m≈3.54 g
Therefore, the theoretical yield of carbon dioxide, CO2, in this reaction is
approximately 3.54 grams.
Question 29
Question
Calculate the mass of ammonium nitrate that can be produced from the reaction
of 50.0 g of ammonia gas (NH3) and 75.0 g of nitric acid (HNO3) according to
the following balanced chemical equation:
NH3+HNO3→N H4N O3
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Solution
Step 1: Write the balanced chemical equation for the reaction.
NH3+HNO3→N H4N O3
Step 2: Calculate the number of moles of each reactant. Given: - Mass
of ammonia gas (NH3) = 50.0 g - Molar mass of ammonia gas (NH3) = 17.03
g/mol - Mass of nitric acid (HNO3) = 75.0 g - Molar mass of nitric acid (HNO3)
= 63.01 g/mol
Number of moles of ammonia gas:
moles NH3=50.0 g
17.03 g/mol = 2.94 mol
Number of moles of nitric acid:
moles HNO3=75.0 g
63.01 g/mol = 1.19 mol
Step 3: Determine the limiting reactant. From the balanced chemical equa-
tion, we see that 1 mol of NH3reacts with 1 mol of HNO3to produce 1 mol of
NH4NO3. To determine the limiting reactant, we need to compare the ratios of
moles of reactants to stoichiometric coefficients. The ratio of moles of NH3to
HNO3is: 2.94 mol NH3
1.19 mol HNO3
≈2.47
So, HNO3is the limiting reactant.
Step 4: Calculate the theoretical yield of ammonium nitrate. From the
reaction equation, 1 mol of NH4NO3has a molar mass of:
14.01 + 4(1.008) + 14.01 + 3(16.00) = 80.04 g/mol
moles NH4NO3= 1.19 mol HNO3×1 mol NH4NO3
1 mol HNO3= 1.19 mol
The theoretical yield of NH4NO3is:
mass NH4NO3= 1.19 mol ×80.04 g/mol = 95.2 g
Therefore, the mass of ammonium nitrate that can be produced is 95.2 g.
Question 30
Question
A researcher is studying a reaction that produces nitrogen monoxide gas. The
reaction is represented by the following balanced chemical equation:
4NH3(g) + 5O2(g)→4NO(g) + 6H2O(g)
If 5.00 moles of ammonia (NH3) react with excess oxygen (O2), how many moles
of water (H2O) will be produced?
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Solution
Step 1: Calculate the moles of water produced by using a mole ratio from the
balanced chemical equation.
Moles of NH3: Moles of H2O=4:6
Step 2: Determine the moles of water produced.
Moles of H2O = 5.00 mol NH3
4×6
1=5.00 ×6
4= 7.50 mol H2O
Therefore, 7.50 moles of water (H2O) will be produced when 5.00 moles of
ammonia (NH3) react.
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