Page 1 of 5
Date: December 6, 2011
By: Swati Kulkarni
Solutions Stoichiometry Worksheet
When performing any kind of stoichiometry calculations by any method, follow
these general steps:
1) Write a balanced equation for the reaction, to obtain MOLE ratios
2) Convert the given value (e.g mass) to an amount in moles using the
appropriate conversion factor
3) Convert the given amount in moles to the required amount in moles, using the
mole ratio from the balanced reaction
4) Convert the required amount in moles to the required value using the
appropriate conversion factor.
Exercises: [ Thinking/ Problem Solving ]
1) Many popular chemical fertilizers include ammonium hydrogen phosphate. This
compound is made commercially by reacting concentrated aqueous solutions of
ammonia and phosphoric acid. What volume of 14.8 mol/L NH3 (aq) would be needed
to react completely with each 1.00 kL of 12.9 mol/L H3PO4 (aq) to produce fertilizer
in a commercial operation?
Step 1: (2 marks)
2NH3 (aq) + H3PO4(aq) → (NH4)2HPO4
2 moles of NH3 produce 1 mole of (NH4)2HPO4
Given: C of NH3 = 14.8 mol/L, C of H3PO4 = 12.9 mol/L, V of H3PO4 = 1.00 kL
Step 2: (2 marks)
To find the number of moles of (NH4)2HPO4 use formula C = n/V
 n = C x V → n of H3PO4 = 1.00 kL x 12.9 mol/L = 12.9 k mol
Step 3: (2 marks)
Using the mole ratio of NH3 to H3PO4 we can find the number of moles of NH3
2 : 1
n of NH3 = 12.9k mol x 2/1 = 25.8 k mol
Step 3: (2 marks)
We can then find volume of NH3 Using V = n/C
Page 2 of 5
Date: December 6, 2011
By: Swati Kulkarni
V of NH3 = 25.8 k mol / 14.8 mol/L and  V of NH3 = 1.74 kL
The volume of NH3 needed would be 1.74 kL
(1 mark)
2) Chemical technologists work in the laboratories of chemical industries. One of
their jobs is to monitor the concentrations of solutions in the process stream. For
example, sulphuric acid is a reactant in the production of sulfates (e.g. ammonium
sulfate in fertilizer plants. A technician needs to determine the concentration of
the sulphuric acid solution. In the experiment, a 10.00 mL sample of sulphuric acid
reacts completely with 15.9 mL of 0.150 mol/L potassium hydroxide solution.
Calculate the molar concentration of the sulphuric acid.
Step 1: (2 marks)
2KOH + H2SO4 (aq) → K2SO4 (aq) + H2O (l)
Given: V of H2SO4 = 10.00 mL, V of KOH = 15.9 mL, C of KOH = 0.150 mol/L
Step 2: (4 marks)
C of KOH = n of KOH/V of KOH
 0.150 mol/L = n of KOH/15.9 mL
 n of KOH = 0.150 mol/Lx 15.9 mL
 n of KOH = 2.39 m mol
Step 3: (3 mark)
As 2 moles of KOH react with 1 mole of H2SO4
 n of H2SO4 = 2.39 m mol x ½
 n of H2SO4 = 1.19 m mol
Step 4: (4 marks)
To find C of H2SO4
C of H2SO4 = n of H2SO4/ V of H2SO4
Page 3 of 5
Date: December 6, 2011
By: Swati Kulkarni
 n of H2SO4 = C of H2SO4/ V of H2SO4
 n of H2SO4 = 1.19 m mol / 10.00 mL
 n of H2SO4 = 0.119 mol/L
The molar concentration of H2SO4 would be 0.119mol/L
(1 mark)
3) What is the mass of precipitate produced by the reaction of 20.0 mL of a 2.50
mol/L stock solution of sodium hydroxide with an excess of zinc chloride solution?
Step 1: (2 marks)
2 NaOH (aq) + ZnCl2 (aq)  Zn(OH)2(s) + 2NaCl (aq)
Given: V of NaOH = 20.0 mL of NaOH = 0.0200 L of NaOH
C of NaOH = 2.50 mol/L
Step 2: (3 marks)
Using C of NaOH = n of NaOH /V of NaOH
n of NaOH = C of NaOH x V of NaOH
 n of NaOH = 2.50 mol/L x 0.0200 L
 n of NaOH = 0.05 mol
Step 3: (3 marks)
2 moles of NaOH produce 1 mole of Zn(OH)2
 n of Zn(OH)2 = 0.05/2 = 0.025 mol
To find # g Zn(OH)2 Use n = m/M
 0.025 mol x 99.424 g/mol = 2.49 g
 The mass of precipitate (zinc hydroxide) produced = 2.49 g
Page 4 of 5
Date: December 6, 2011
By: Swati Kulkarni
The mass of zinc hydroxide produced is = 2.49 g
(1 mark)
4) Lucy and Alena wish to precipitate all of the lead (II) ions from 2.00 L of
solution containing, among other substances, 0.34 M of lead nitrate. The purpose of
this reaction is to make the filtrate solution non-toxic. If Lucy and Alena intend to
precipitate lead (II) sulphate, suggest an appropriate solute and calculate the
appropriate concentration and volume of the reacting solution.
Step 1: (2 marks)
An appropriate solute would be K2SO4 (aq)
The reaction is:
Pb(NO3)2(aq) + K2SO4 (aq) → PbSO4 (s) + 2KNO3(aq)
1:1
Step 2: (2 marks)
1:2
Given: 0.34 M of lead nitrate that is 0.34 mols/1L
The solution is 2.00 L and hence
Number of moles n = 0.34 mols/1L x 2.00 L = 0.68 moles of lead nitrate
Step 3: (4 marks)
1 mole of lead nitrate reacts with 1 mole of potassium sulphate and hence number of
moles of potassium sulphate is 0.68 moles.
To determine the mass of potassium sulphate use the formula, m = n x molar mass
m=0.68 moles x 174.26 g/moles
 The mass of potassium sulphate = 118.5 g
Concentration = (mass of solute in g) / volume solution in L C = 118.5 g/2L
 C = 59.25 g/L and volume = 2L
Page 5 of 5
Date: December 6, 2011
By: Swati Kulkarni
5) What is the percent yield of the precipitate if 20.0 mL of 0.210 M sodium
sulphide are placed in the same beaker as an excess quantity of aluminium nitrate
and 0. 190 g of precipitate are measured? Suggest some reasons why 100%
recovery of the precipitate might not be achieved.
Step 1: (2 marks)
3 Na2S (aq) + 2 Al(NO3)3(aq) -----> 6 NaNO3(aq) + Al2S3(s)
A
B
Step 2: (3 marks)
 g B = 0.210molA/L x 0.0200 L x 1 mol B/ 3molsA x 150.14gB/1molB =0.210 g
 From the reaction 0.210 g of precipitate (aluminum sulphide) should be formed.
Experimentally 0.190 g of precipitate are measured
 Percent yield = actual yield ÷ theoretical yield x 100
 Percent yield = 0.190 ÷ 0.210 x 100 = 90.47%
The percent yield of the precipitate is = 90.47%
(1 mark)
Suggest some reasons why 100% recovery of the precipitate might not be
achieved. (1 mark)
As aluminium nitrate is in excess quantity, sodium sulphide is the limiting reagent.
If the reaction is incomplete it will affect the percent yield. Also, due to the loss in
the process or inefficiency of the chemical reaction it is not possible to recover
100 % precipitate.
This Problem I will use for: Assessment Of Learning