Stoichiometry Practice Problems
1. What is the percent composition by mass of pentyl acetate, CH3CO2C5H11, the
compound responsible for the odor of bananas?
64.6 % Carbon, 24.6 % Oxygen, 10.8 % Hydrogen
2. For a 10.0 g sample of glucose, C6H12O6, calculate the following:
a) the number of moles of glucose molecules.
0.055 moles of glucose molecules
b) the number of glucose molecules.
3.34x1022 molecules of glucose
c) the number of carbon atoms.
2.00x1023 carbon atoms
d) the moles of carbon atoms.
0.333 moles of carbon atoms
e) the mass of carbon in the sample.
4.0 g of carbon
3. A 4.000 g sample of a solid mixture is known to contain only sodium chloride
and potassium chloride. The sample is dissolved in water and excess silver
nitrate is added resulting in the formation of a white precipitate. When the
precipitate is filtered from the solution and dried it weighs 8.5904 g.
a) What is the precipitate and write the balanced net ionic equation for the
reaction that is occurring to form the precipitate.
The precipitate is AgCl(s)
The net ionic equation is: Ag+(aq) + Cl-(aq) --> AgCl(s)
b) Calculate the percentage by mass of NaCl and KCl in the original
sample.
The Cl- in the AgCl precipitate comes from two sources, either the
KCl or the NaCl in the original sample. The total moles of Cl- in the
sample which contains NaCl and KCl is equal to the number of moles
of AgCl which can be determined as follows: Moles AgCl = moles Cl= (8.5904g AgCl)(1 mol/143.3 g AgCl) = 0.05994 mol ClIf we let x = moles of Cl- from NaCl and y = moles of Cl- from KCl then:
x + y = 0.05994 mol and
(58.44 g/mol)x + (74.55 g/mol)y = 4.000 g
then: y = 0.05994 - x substituting this into the above equation gives:
(58.44 g/mol)x + (74.55 g/mol)( 0.05994 - x) = 4.000 g
Solving for x yields x = 0.02911 mol Cl- from NaCl
Using the result for x to get y yields y = 0.3083 mol of Cl- from KCl
The mass of NaCl is (0.02911 mol)(58.44 g/mol) = 1.701 g
The mass of KCl is (0.03083 mol)(74.55 g/mol) = 2.299 g
so the original compound is 42.5% NaCl and 57.5% KCl
c) If you had a 0.100 M AgNO3 solution for use in the precipitation
reaction, how many milliliters would be required to completely react with
all of the original sample?
We would need 0.05994 moles of Ag+ and thus 0.5994 moles of
AgNO3 Volume = (0.05944 mol AgNO3)(1 L/0.100 mol AgNO3) = 0.5994
L = 599.4 L
4. With the news that calcium deficiencies in many people’s diets may contribute
to the development of osteoporosis (bone weakening), the use of dietary calcium
supplements has become increasingly common. Many of these calcium
supplements consist of nothing more than calcium carbonate, which is also in
Tums, and serves the dual purpose there of
providing a calcium supplement and neutralizing stomach acid! When a calcium
carbonate tablet is ingested, it dissolves by reaction with stomach acid, which
contains hydrochloric acid. The UNBALANCED chemical equation is:
CaCO3 (s) + HCl (aq) ----> CaCl2 (aq) + H2O (l) + CO2 (g)
a) Write the balanced net ionic equation.
CaCO3 (s) + 2 H+ (aq) ----> Ca+2 (aq) + H2O (l) + CO2 (g)
b) How many moles of CaCO3 are there in a tablet contains 500 mg of
CaCO3.
5x10-3 mol of CaCO3
c) What volume of 1.25 M HCl is required to react with this tablet?
8.00x10-3 L or 8.00 mL
5. When a sample of a compound in the vitamin D family with a molecular weight
of 399 g/mol was burned in a combustion analysis, 5.982 mg of the compound
gave 5.04 mg of carbon, 0.699 mg of hydrogen, and 0.242 mg of oxygen.
a) What is the percent composition by mass of C, H, and O in the
compound?
84.3 % carbon, 11.7 % Hydrogen, 4.1 % Oxygen
b) What is the empirical formula of the compound?
C28H46O
c) What is the molecular formula of the compound?
C28H46O
6. Cisplatin [Pt(NH3)2Cl2], a compound used in cancer treatment, is prepared by
reaction of ammonia with potassium tetrachloroplatinate: K2PtCl4 + 2NH3 --> 2KCl
+ Pt(NH3)2Cl2
a) What is the theoretical yield of cisplatin (in grams) that can be formed
from 55.8 g of K2PtCl4 and 35.6 g of NH3.
40.3 g of cisplatin
b) Which reactant is limiting?
K2PtCl4
c) If 34.8 g of cisplatin are actually formed, what is the percentage yield?
86.4 %
d) If you wanted to make up 500 mL of a 1.00 x 10-3 M solution of cisplatin,
how many grams of cisplatin would you need?
0.150 g or 150 mg of cisplatin
7. Fumaric acid, which occurs in many plants, contains, by mass, 41.4% carbon,
3.47% hydrogen, and 55.1% oxygen. A 0.050-mole sample of this compound
weighs 5.80 g. What is the molecular formula of fumaric acid?
C4H4O4
8. A 10.00 mL sample of sulfuric acid from an automobile battery requires 35.08
mL of a 2.12 M sodium hydroxide solution for complete neutralization. What is
the molarity of the sulfuric acid? The reaction is:
H2SO4(aq) + 2NaOH(aq) --> 2H2O(l) + Na2SO4(aq)
3.72 M H2SO4
9. Hydrogen and chloride react to yield hydrogen chloride:
H2(g) + Cl2(g) --> 2HCl(g)
a) How many grams of HCl can be produced from the reaction of 3.56 g of
H2 and 8.94 g of Cl2?
9.19 g of HCl
b) How many liters of HCl is this at standard temperature and pressure
(assume ideal gas behavior)?
5.64 L of HCl
c) Which reactant is limiting?
Cl2
d) If 4.32 L of HCl is actually produced, what is the percentage yield of the
reaction (assume ideal gas bejavior)?
76.6 %