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:
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?
Answer:
step 1: (TI: 2marks)
2NH3(aq) + H3PO4(aq)
(NH4)2HPO4(aq)
v=?
v=1.00kL
C=14.8mol/L
C=12.9mol/L
step 2: (TI: 2 marks)
to find the number of moles of H3PO4, we use the formula C=n/v Molar Conc. = Amount of solute
so, n=v x C
nH3PO4=1.00kL x 12.9mol/L =12.9kmol
Volume of Solution
step 3: (TI:2marks)
using the mole ratio of NH3 : H3PO4
2 : 1
nNH3 = 12.9kmol x 2/1 = 25.8kmol
step 4:(TI 2marks) again using C=n/v, to find the volume of NH3
v=n/C
vNH3=25.8kmol
14.8mol/L
v=1.74kL
Therefore the volume of NH3 needed would be 1.75kL. (C:1mark)
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.
Answer:
step 1: H2SO4(aq) + 2KOH
2H2O(l) + K2SO4(aq) (TI 2marks)
v=10.00mL
v=15.9mL
C=?
C=0.150mol/L
step 2: CKOH =nKOH/vKOH (TI 1mark)
therefore: nKOH =vKOH x CKOH
=15.9mL x 0.150mol/L
=2.39mmol
step 3: nH2SO4 =2.39mmol x 1 (TI 1mark)
2
nH2SO4 =1.19mmol
step 4: to find C of H2SO4 (TI 1mark)
C H2SO4 = n/v = 1.19mmol
10.00mL
= 0.119mol/L
Therefore the molar concentration of the sulfuric acid is 0.119mol/L. (C: 1mark)
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?
Answer:
Step 1: (2 marks)
2 NaOH (aq) + ZnCl2 (aq)
Zn(OH)2(s) + 2NaCl (aq) (since NaCl easily dissolves, the
precipitate must be Zn(OH)2)
Given: V of NaOH = 20.0 mL of NaOH Convert to L = 0.0200 L of NaOH
C of NaOH = 2.50 mol/L
Step 2: (3 marks) Use C = n/v to find amount of solute of NaOH
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) Mole ratio of NaOH: Zn(OH)2
2 moles of NaOH produce 1 mole of Zn(OH)2
n of Zn(OH)2 = 0.05/2 = 0.025 mol
To find mass in grams of Zn(OH)2 use molar mass of Zn(OH)2 x Amount (in mol)
To determine the molar mass of Zn(OH)2, we add the atomic mass of each
Zn=65.41
2O=16*2=32
2H=1.01*2=2.02 Summed up= 99.43
Therefore, 0.025 mol x 99.424 g/mol = 2.49 g
The mass of precipitate (zinc hydroxide) produced = 2.49 g (C: 1mark)
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: (TI 2 marks)
An appropriate solute would be Na2SO4 (aq)
The reaction is:
Pb(NO3)2(aq) + Na2SO4 (aq) → PbSO4 (s) + 2NaNO3(aq)
1:1
Step 2: (TI 2 marks)
1:2
Given: 0.34 M of lead nitrate
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: (TI 4 marks) Mole Ratio
1 mole of lead nitrate reacts with 1 mole of sodium sulphate and hence number of moles of
sodium sulphate is 0.68 moles
To determine the mass of sodium sulphate use the formula, m = n(#0f moles) x molar mass
m=0.68 moles x 142.05 g/mole
The mass of sodium sulphate = 96.59 g
Concentration = (mass of solute in g) / volume solution in L
C = 96.59 g/2L
C = 48.3 g/L
Therefore, concentration of reacting solution is 48.3g/L and Volume is 2L (C: 1mark)
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.
Answer:
Step 1: (TI 2marks)
3 Na2S (aq) + 2 Al(NO3)3(aq) ---- 6 NaNO3(aq) + Al2S3(s)
Step 2: (2marks)
Find out moles of Na2S n = C X V
0.210 mol/L x 0.02L
= 4.2 x 10-3mols
Find out moles of Al2S3 using molar ratios:
3 mols of Na2S = 4.2 x 10-3mols
1 mol of Al2S3 Al2S3 mols
-3
Mols of Al2S3 = 1.4 x 10
Step 3: (TI: 2marks) Calculate mass of Al2S3(theoretical yield) using molecular weight of
150.17g/mol
m = 150.17g x 1.4 x 10-3
mol
mass of Al2S3 = 0.210g Therefore percent yield = Actual yield
Theoretical yield
= 0.190g x 100%
0.210g
= 90.48%
Percent yield of precipitate is 90.48%
Suggest some reasons why 100% recovery of the precipitate might not be achieved. (C: 2
mark)
The reaction might be incomplete resulting in less than 100% recovery of the precipitate due to
temperature of solution which might dissolve precipitate and force the reaction to go in the
opposite direction. If there are impurities in solution, it may use up some of the limiting reagent.
The reaction might be less efficient
Assessment of Learning Total: C: /6 TI: /35