Stoichiometry Involving Solutions Worksheet Answers
1. Calculate the number of mL of 2.00 M HNO3 solution required to react with 216 grams of Ag
according to the equation.
3 Ag(s) + 4 HNO3(aq) ---------> 3 AgNO3(aq) + NO(g) + 2 H2O(l)
Set up Stoichiometry Table
grams
MM
moles
Ag
216
108
2
HNO3
2.7
M = moles/L so 2.0 = 2.7/x, x =1.35L or 1350 mL
2. Calculate in mL the volume of 0.500 M NaOH required to react with 3.0 grams of acetic acid.
The equation is:
NaOH(aq) + HC2H3O2(aq) -------> NaC2H3O2(aq) + H2O(l)
Set up Stoichiometry Table
NaOH
HC2H3O2
grams
3g
MM
60g/mol
moles
0.05 moles 0.05 moles
M = moles / L, so 0.5 = 0.05 / x, x = 0.1L or 100 ml.
3. Calculate the number of grams of AgCl formed when 0.200 L of 0.200 M AgNO3 reacts with
an excess of CaCl2. The equation is:
2 AgNO3(aq) + CaCl2(aq) -------> 2 AgCl(s) + Ca(NO3)2(aq)
Calculate moles of AgNO3 used
M = mol/L, 0.2 = x / 0.2L, x = 0.04 moles
Use the stoichiomety table
AgNO3
Grams
MM
Moles
0.04 moles
5.74 g of AgCl can be produced
CaCl2

AgCl
5.74g
143.4g/mol
0.04 moles
Ca(NO3)2
4. Calculate the mass of AgCl formed when an excess of 0.100 M solution of NaCl is added to
0.100 L of 0.200 M AgNO3.
Calculate the moles AgNO3 used.
M = mol/L, 0.2 = x / 0.1L, x = 0.02 moles
AgNO3
Grams
MM
Moles
2.89g of AgCl.
CaCl2

0.02 moles
AgCl
2.89g
143.4g/mol
0.02 moles
Ca(NO3)2
5. Calculate:
a) the mass of BaSO4 formed when excess 0.200 M Na2SO4 solution is added to 0.500 L of 0.500
M BaCl2 solution, and
Write the balanced equation
Na2SO4 + BaCl2  2NaCl + BaSO4
Calculate the moles of BaCl2.
M = mol/L, 0.5 = x / 0.5L, x = 0.25 mol
Set up stoichiometry table.
Na2SO4
BaCl2
Grams
MM
Moles
0.25 moles
58.4g of BaSO4 should be produced

NaCl
BaSO4
58.4g
233.4 g/mol
0.25 moles
b) the minimum volume of the Na2SO4 solution needed to precipitate the Ba2+ ions from the
BaCl2 solution.

Na2SO4
BaCl2
NaCl
BaSO4
Grams
35.5g
58.4g
MM
142.1 g/mol
233.4 g/mol
Moles
0.25 moles
0.25 moles
0.25 moles
Since it is a 1:1 ratio 0.25 moles of Na2SO4 are needed use the Molarity formula to determine
the # of Liters needed.
M = mol/L, 0.2 = 0.25/x, x = 1.25 L or 1250 mL. There would be 35.5 g of Na2SO4 in the 1250
ml.
6. A sample of impure sodium chloride weighing 1.00 grams is dissolved in water and
completely reacted with silver nitrate solution. The dried precipitate of AgCl has a mass of 1.48
grams . Calculate the percentage of NaCl in the original impure sample.
NaCl
0.585g
58.5 g/mol
0.010 moles
Grams
MM
Moles

AgNO3
AgCl
1.48 g
143.4 g/mol
0.010 moles
NaNO3
There were 0.585 g of NaCl in the 1 g sample so the % NaCl is 58.5%
7. To neutralize the acid in 10.0 mL of 18.0 M H2SO4 that was accidentally spilled on a
laboratory bench top, solid sodium bicarbonate was used. The container of sodium bicarbonate
was known to weigh 155.0 g before this use and out of curiosity its mass was measured as 144.5
g afterwards. The reaction that neutralizes sulphuric acid this way is as follows.
H2SO4 + 2 NaHCO3 --------> Na2SO4 + 2 CO2 + 2 H2O
Calculate the amount of NaHCO3 used
Mass of container before =
155.0 g
Mass of container after
=
144.5 g
Mass of NaHCO3
=
10.5g
Was sufficient sodium bicarbonate used? Calculate the limiting reactant and the maximum yield
in grams of sodium sulphate.
H2SO4
grams
MM
moles
Moles
neutralized
NaHCO3
10.5g
84.0g/mol
0.125 mol

Na2SO4
CO2
H2O
0.0625
moles
Based on stoichiometry we can neutralize 0.0625 moles of H2SO4, use the molarity formula to
see how many moles of H2SO4 were present.
M = mol/L, 18 = mol / 0.010L, mol = 0.18 so since 0.18mol > 0.0625 moles we did not
neutralize all the acid.
8. Barium nitrate and potassium sulphate solutions react and form a precipitate. What is the
precipitate? How many mL of 0.40 M Ba(NO3)2 solution are required to precipitate completely
the sulphate ions in 25 mL of 0.80 M K2SO4 solution?
Write the balanced reaction
Ba(NO3)2 + K2SO4  2KNO3 + BaSO4.
Using solubility rules BaSO4 is the precipitate.
Calculate moles of K2SO4 present
M = mol/L, 0.8 = mol / 0.025L, mol = 0.02 mol.
Since the K2SO4 and Ba(NO3)2 are in a 1:1 ratio the moles needed of Ba(NO3)2 is 0.02 moles.
Use the molarity formula to calculate the volume of Ba(NO3)2 needed.
M = mol/L, 0.4 = 0.02 mol / L, L = 0.05L or 0 mL.
9. What mass of silver chloride can be precipitated from a silver nitrate solution by 200 mL of a
solution of 0.50 M CaCl2?
Calculate moles of CaCl2 used
M = mol/L, 0.5 = mol/0.2L, mol = 0.01
Write the balanced equation
CaCl2 + 2AgNO3  Ca(NO3)2 + 2AgCl
Use Stoichiometry to solve the problem
CaCl2
grams
MM
moles
0.01 moles
2.87g of AgCl can be produced.
AgNO3

Ca(NO3)2
AgCl
2.87 g
143.4 g/mol
0.02 moles