YISHUN INNOVA JUNIOR COLLEGE

2019 JC1 H2 CHEMISTRY

TUTORIAL 2
SOLUTION STOICHIOMETRY

A Balanced Equations and Stoichiometric Mole Ratio

[Refer to lecture notes pages 3 – 7]

1. Calculate the volume of SO2, measured at s.t.p., required to react with 6.0 g of O2, to produce the
maximum mass of SO3, by the following reaction: 2SO2 + O2  2SO3.
(8.51 dm3)

Amount of O2 = 6.0  2(16.0) = 0.1875 mol

Mole ratio of SO2 : O2 = 2 : 1
Amount of SO2 = 2  0.1875 = 0.375 mol
Volume of SO2 = 0.375  22.7 = 8.51 dm3

2. Chromium can be extracted from its chromium(III) oxide by the Thermit reaction,
2Al + Cr2O3  2Cr + Al2O3
Calculate the percentage yield when 180 g of chromium are obtained from a reaction between
100 g of aluminium and 400 g of chromium(III) oxide.
(93.5%)

Amount of Al = 100  27.0 = 3.7037 mol

Amount of Cr2O3 = 400  152 = 2.6315 mol
Mole ratio of Al : Cr2O3 = 2 : 1
2.6315 mol of Cr2O3 would react completely with 5.2630 mol of Al.
Since only 3.7037 mol of Al is available, Al is the limiting reagent.

Mole ratio of Al : Cr = 1 : 1
Amount of Cr = 1  3.7037 = 3.7037 mol
Theoretical yield = 3.7037  52.0 = 192.59 g
180
Percentage yield =  100% = 93.5%
192.59
Tutorial 2: Solution Stoichiometry

3. N14/P3/Q2(e)
After the reduction is finished, excess LiAlH4 is destroyed as follows:
 Water is added; hydrogen is evolved.
 Dilute acid is added to dissolve the mixture of hydroxides.

(a) Suggest equations for the reactions that occur during the two processes above.

(b) Calculate the volume of hydrogen at room temperature and pressure that would be evolved
if 2.00 g of LiAlH4 were decomposed in this way.
(5.07 dm3)

(a) LiAlH4 + 4H2O  LiOH + Al(OH)3 + 4H2

LiOH + HCl  LiCl + H2O
Al(OH)3 + 3HCl  AlCl3 + 3H2O

(b) Amount of LiAlH4 = 2.00  37.9 = 0.052770 mol

Mole ratio of LiAlH4 : H2 = 1 : 4
Amount of H2 = 4  0.052770 = 0.21108 mol
Volume of H2 = 0.21108  24.0 = 5.07 dm3

4. In an experiment, 0.130 g of zinc, is added to 100 cm3 of 0.0500 mol dm−3 copper(II) sulfate,
CuSO4, solution.

(a) Write an ionic equation, including state symbols, for the reaction between zinc and
copper(II) sulfate.

(b) Calculate the mass of copper that will be precipitated.

(0.126 g)

(a) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

(b) Amount of Zn = 0.130  65.4 = 1.9877  103 mol

Amount of Cu2+ = 0.0500  (100  103) = 5.00  103 mol
Mole ratio of Zn : Cu2+ = 1 : 1
Since 5.00  10−3 mol of CuSO4 requires 5.00  103 mol of Zn, and only 1.9877  10−3 mol of Zn
present, hence Zn is limiting reagent.

Mole ratio of Zn : Cu = 1 : 1
Amount of Cu = 1  (1.9877  103) = 1.9877  103 mol
Mass of Cu = (1.9877  103)  63.5 = 0.126 g

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Tutorial 2: Solution Stoichiometry

B Concentration of Solutions
[Refer to lecture notes pages 7  14]

5. What volumes of 0.0500 mol dm−3 HCl and 0.0100 mol dm−3 HCl must be mixed to give 2.0 dm3
of 0.0200 mol dm−3 HCl?
(0.500 dm3 and 1.50 dm3)

Let the volume of the 0.0500 mol dm-3 HCl required be x dm3,
and the volume of the 0.0100 mol dm-3 HCl required be (2.0 − x) dm3.
Amount of HCl in the final solution = c1V1 + c2V2
0.0200  2 = 0.0500x + 0.0100(2.0 – x)
x = 0.500
Volume of the 0.0500 mol dm-3 HCl required is 0.500 dm3,
Volume of the 0.0100 mol dm-3 HCl required is 2.0 − 0.500 = 1.50 dm3

6. AJC/2014/P2/Q2(a)
Blood alcohol concentration (BAC) is a good measure of the extent to which the activity of the
central nervous system is depressed. It is usually defined as follows:

BAC = mg of ethanol per 100 cm3 of blood

At present in Singapore, the legal limit to drive is 80 mg of ethanol per 100 cm 3 of blood. If the
blood of the driver is found to exceed this legal limit, he will be charged with drink–driving. Ethanol
is sufficiently volatile to pass from the blood into the air in the lungs and the following equilibrium
is set up.

C2H5OH (blood) C2H5OH (g)

As the ethanol concentration in the breath is related to that in the blood, a traffic police officer can
test the driver’s breath using a device called a breathalyser instead of drawing a driver’s blood to
test his ethanol level. At the temperature of the human body, the ratio of breath ethanol to blood
ethanol is 2100 : 1. This ratio states that 2100 cm3 of exhaled air (breath) contains the same
amount of ethanol as in 1 cm3 of blood.

In one type of breathalyser, the exhaled air is passed through a solution of acidified potassium
dichromate which oxidises the ethanol to ethanoic acid in an electrochemical cell. The electrical
current generated by the oxidation can be used to estimate the ethanol content of blood.

What will be the breath alcohol concentration, in mol cm–3, which corresponds to the legal limit for
BAC?

(8.28  10−9 mol cm3)

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Tutorial 2: Solution Stoichiometry

Legal limit
= 80 mg per 100 cm3 blood
= 0.80 mg per cm3 blood
= 0.80mg per 2100 cm3 breath
0.80
= = 0.00038095 mg per cm3 breath
2100

0.00038095  10 3
= = 8.28 x 10–9 mol cm–3
46.0

C Titration
[Refer to lecture notes pages 15  18]

7. A dibasic acid is one in which one mole of the acid dissociates to give two moles of H+.
Phosphoric(V) acid, H3PO4, is a dibasic acid.

(a) Write a balanced equation for the reaction between H3PO4 and NaOH.

(b) Calculate the volume of 0.050 mol dm−3 phosphoric(V) acid required to neutralise
20.0 cm3 of 0.100 mol dm−3 sodium hydroxide.
(0.0200 dm3)

(a) H3PO4 + 2NaOH Na2HPO4 + 2H2O

(b) Amount of NaOH = 0.100  (20.0  103) = 2.00  103 mol

Mole ratio of H3PO4 :NaOH = 1 : 2
Amount of H3PO4 = 0.5  (2.00  103) = 1.00  103 mol
Volume of H3PO4 = (1.00  103)  0.050 = 0.0200 dm3

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Tutorial 2: Solution Stoichiometry

8. Distilled water was added to 24.0 cm3 of sulfuric acid to give a total volume of 250 cm3. When
25.0 cm3 of this solution is titrated, it required 22.80 cm3 of 1.325 g dm−3 of NaOH. Calculate the
original concentration of sulfuric acid.
(0.157 mol dm3)

H2SO4 + 2NaOH Na2SO4 + 2H2O

Concentration of NaOH in mol dm3 = 1.325  40.0 = 0.033125 mol dm3

Amount of NaOH 0.033125  (22.80  103) = 7.5525  104 mol
Mole ratio of H2SO4 :NaOH = 1 : 2
Amount of H2SO4 in 25.0 cm3 = 0.5  (7.5525  104) = 3.7762  104 mol
250
Amount of H2SO4 in 250 cm3 =  (3.7762  104) = 3.7762  103 mol
25

Since 24.0 cm3 H2SO4 was diluted to 250 cm3, the amount of H2SO4 in both are the same.
Amount of H2SO4 in 24.0 cm3 = 3.7762  103 mol
Concentration of H2SO4 = (3.7762  103)  (24.0  103) = 0.157 mol dm3

9. When iodic(V) acid, HIO3, reacts with an aqueous solution of sulfur dioxide, sulfuric acid and
iodine are produced. In an experiment, 0.100 g of iodic(V) acid was reacted with excess sulfur
dioxide and iodine formed were removed by heating. The sulfuric acid produced required
28.40 cm3 of 0.100 mol dm−3 NaOH solution for complete neutralisation. Determine the equation
for the reaction between iodic(V) acid and sulfur dioxide.

H2SO4 + 2NaOH Na2SO4 + 2H2O

Amount of NaOH = 0.100  (28.40  103) = 2.84  103 mol

Mole ratio of H2SO4 :NaOH = 1 : 2
Amount of H2SO4 = 0.5  (2.84  103) = 1.42  103 mol
Amount of HIO3 = 0.100  176.0 = 5.6818  104 mol

0.00056818 0.00142
Mole ratio of HIO3 : H2SO4 = 5.6818  104 : 1.42  103 = : = 1 : 2.4992
0.00056818 0.00056818
=2:5

2HIO3 + 5SO2 + 4H2O 5H2SO4 + I2

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Tutorial 2: Solution Stoichiometry

10. Sulfur dichloride oxide, SO2Cl2, is hydrolysed by water to sulfuric acid and hydrochloric acid. A
sample of SO2Cl2 was dissolved in sufficient water to produce 150.0 cm3 of solution, and 25.0
cm3 of the resultant solution required 15.00 cm3 of 0.200 mol dm−3 sodium hydroxide for
neutralisation.

(a) Write a balanced equation for the hydrolysis of SO2Cl2.

(b) Calculate the mass of SO2Cl2 in the sample.

(0.608 g)

(a) SO2Cl2 + 2H2O H2SO4 + 2HCl

(b) H+ + OH H2O

Amount of NaOH = 0.200  (15.00  103) = 3.00  103 mol
Mole ratio of H+ : OH = 1 : 1
Amount of H+ in 25.0 cm3 = 1  (3.00  103) = 3.00  103 mol
150
Amount of H+ in 150.0 cm3 =  (3.00  103) = 0.0180 mol
25

Mole ratio of H+ : SO2Cl2 = 1 : 4

Amount of SO2Cl2 = 0.25  0.0180 = 4.50  103 mol
Mass of SO2Cl2 = (4.50  103)  135.1 = 0.608 g

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Tutorial 2: Solution Stoichiometry

D Back Titration
[Refer to lecture notes pages 19  24]

11. 1.65 g of an impure sample of barium hydroxide, Ba(OH)2 was allowed to react with 100 cm3 of
0.200 mol dm−3 hydrochloric acid. It was found that the resultant solution required 10.90 cm 3 of
0.250 mol dm−3 NaOH for neutralisation. Calculate the percentage purity of the sample of barium
hydroxide.
(89.7%)

HCl + NaOH  NaCl + H2O

Amount of NaOH = 0.250  (10.90  103) = 2.725  103 mol

Mole ratio of HCl : NaOH = 1 : 1
Amount of excess HCl = 1  (2.725  103) = 2.725  103 mol

Amount of HCl added = 0.200  (100  103) = 0.0200 mol

Amount of HCl reacted = 0.0200  (2.725  103) = 0.017275 mol

2HCl + Ba(OH)2  BaCl2 + 2H2O

Mole ratio of HCl : Ba(OH)2 = 2 : 1

Amount of Ba(OH)2 = 0.5  0.017275 = 8.6375  103 mol
Mass of Ba(OH)2 = (8.6375  103)  171.3 = 1.4796 g
1.4796
Percentage purity of Ba(OH)2 =  100% = 89.7%
1.65

12. 1.70 g of an impure sample of ammonium sulfate contaminated with sodium sulfate, was dissolved
in sufficient water to form 250 cm3 of solution. 25.0 cm3 of this solution was boiled with 50.0 cm3
of 0.100 moldm−3 sodium hydroxide solution, and the excess alkali was found to require
12.40 cm3 of 0.200 mol dm−3 HCl solution. Calculate the percentage by mass of ammonium sulfate
in the sample.
(97.9%)

HCl + NaOH  NaCl + H2O

Amount of HCl = 0.200  (12.40  103) = 2.48  103 mol

Mole ratio of HCl : NaOH = 1 : 1
Amount of excess NaOH = 1  (2.48  103) = 2.48  103 mol

Amount of NaOH added = 0.100  (50  103) = 5.00  103 mol

Amount of NaOH reacted = (5.00  103)  (2.48  103) = 2.52  103 mol

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Tutorial 2: Solution Stoichiometry

(NH4)2SO4 + 2NaOH  2NH3 + Na2SO4 + 2H2O

Mole ratio of (NH4)2SO4 : NaOH = 1 : 2

Amount of (NH4)2SO4 in 25.0 cm3 = 0.5  (2.52  103) = 1.26  103 mol
250
Amount of (NH4)2SO4 in 250 cm3 =  (1.26  103) = 0.0126 mol
25

Mass of (NH4)2SO4 = 0.0126  132.1 = 1.6644 g

1.6644
Percentage by mass of (NH4)2SO4 =  100% = 97.9%
1.70

13. 1.350 g of an insoluble carbonate, MCO3 was dissolved in 250 cm3 of a 0.203 mol dm−3
hydrochloric acid solution. The resulting solution was boiled off to remove all the carbon dioxide
produced. 25.0 cm3 of the solution required 24.10 cm3 of 0.100 mol dm−3 sodium hydroxide
solution. Determine the relative atomic mass of M.
(41.3)

HCl + NaOH  NaCl + H2O

Amount of NaOH = 0.100  (24.10  103) = 2.41  103 mol

Mole ratio of HCl : NaOH = 1 : 1
Amount of excess HCl in 25.0 cm3 = 1  (2.41  103) = 2.41  103 mol
250
Amount of excess HCl in 250 cm3 =  (2.41  103) = 0.0241 mol
25

Amount of HCl added = 0.203  (250  103) = 0.05075 mol

Amount of HCl reacted = 0.05075 - 0.0241  = 0.02665 mol

2HCl + MCO3  MCl2 + CO2 + H2O

Mole ratio of HCl : MCO3 = 2 : 1

Amount of MCO3 = 0.5  0.02665 = 0.013325 mol
Molar mass of MCO3 = 1.350  0.013325 = 101.31 g mol1
Molar mass of M = 101.31 – 12.0 – 3(16.0) = 41.3 g mol1
Relative atomic mass of M = 41.3

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Tutorial 2: Solution Stoichiometry

E Additional Questions
14. ACJC/2011/P1/Q1
The hardness present in a water sample due to dissolved calcium ions can be determined by
using ion-exchange column as shown in the diagram.
Ca2+ + 2H+-resin  Ca2+-resin + 2H+
A 50 cm3 sample of a solution containing calcium sulfate was passed through the ion-exchange
resin. The calcium ions in the sample were quantitatively exchanged by hydrogen ions. The
sample collected in the flask required 25 cm3 of 1.0 × 102 mol dm3 potassium hydroxide for
complete neutralisation.

What was the concentration of the calcium sulfate in the original sample?
(2.50 × 103 mol dm3)

Amount of KOH = 0.0100  (25  103) = 2.50  104 mol

Mole ratio of KOH : H+ = 1 : 1
Amount of H+ = 1  (2.50  104) = 2.50  104 mol
Mole ratio of H+ : Ca2+ = 2 : 1
Amount of Ca2+ = 0.5  (2.50  104) = 1.25  104 mol
Concentration of CaSO4 = (1.25  104) ÷ (50  103) = 2.50  103 mol dm3

15. N09/P3/Q3(c)
When sodium is burned in air, a mixture of sodium oxide, Na2O, and sodium peroxide, Na2O2, is
formed. The mixture reacts with water according to the following equations.
Na2O + H2O  2NaOH
Na2O2 + 2H2O  2NaOH + H2O2
The following information will allow you to calculate the relative amounts of the two oxides
produced when sodium is burned.
 The mixture obtained by burning a sample of sodium was dissolved in distilled water and made
up to 100 cm3 to give solution H.
 A 25.0 cm3 portion of solution H was titrated with 0.100 mol dm-3 HCl.
22.5 cm3 of acid was required to reach the end-point.
 The H2O2 content of solution H was found by titration of another 25.0 cm3 of portion with
0.0200 mol dm-3 KMnO4.

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Tutorial 2: Solution Stoichiometry

The following reaction occurs.

2MnO4− + 5H2O2 + 6H+  2Mn2+ + 5O2 + 8H2O
10.0 cm3 of KMnO4 solution was required to reach the end-point.

(a) Using the results of the HCl titration, calculate the total amount in moles of NaOH in
100 cm3 of solution H.
(9.00 × 103 mol)

(b) Using the results of the KMnO4 titration, calculate the amount in moles of H2O2 in 100 cm3
of solution H.
(2.00 × 103 mol)

(c) Hence, calculate the amount in moles of Na2O and Na2O2 formed during the burning of the
sodium sample.
(2.00 × 103 mol; 2.50 × 103 mol)

(a) HCl + NaOH  NaCl + H2O

Amount of HCl = 0.100  (22.5  103) = 2.25  103 mol

Mole ratio of HCl : NaOH = 1 : 1
Amount of NaOH in 25 cm3 = 2.25 × 10−3 mol
Amount of NaOH in 100 cm3 = (2.25 × 10−3) × 4 = 0.00900 mol

(b) Amount of KMnO4 used = 0.0200 × 0.010 = 2.00 × 10−4 mol

Mole ratio of MnO4 : H2O2 = 2 : 5
5
Amount of H2O2 in 25.0 cm3 = 2 × 2.00 × 10−4 = 5.00 × 10−4 mol
Amount of H2O2 in 100 cm3 = (5.00 × 10−4) × 4 = 0.00200 mol

(c) Mole ratio of Na2O2 : H2O2 = 1 : 1

Amount of Na2O2 = 0.00200 mol

Mole ratio of Na2O2 : NaOH = 1 : 2

Mole ratio of Na2O : NaOH = 1 : 2
Amount of NaOH from Na2O2 = 2 × 0.00200 = 0.00400 mol
Amount of NaOH from Na2O = 0.00900 − 0.00400 = 0.00500 mol
Amount of Na2O2 = 0.00500 ÷ 2 = 0.00250 mol

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Tutorial 2: Solution Stoichiometry

16. N10/P3/Q4(f)
Ozone is usually made by passing oxygen gas through a tube between two highly charged
electrical plates.
3O2(g)  2O3(g)
The reaction does not go to completion, so a mixture of the two gases results. The concentration
of O3 in the mixture can be determined by its reaction with aqueous KI.
O3 + 2KI + H2O  I2 + O2 + 2KOH
The iodine formed can be estimated by its reaction with sodium thiosulfate.
2Na2S2O3 + I2  Na2S4O6 + 2NaI
When 500 cm3 of an oxygen/ozone gaseous mixture at s.t.p. was passed into an excess of
aqueous KI, and the iodine titrated, 15.0 cm3 of 0.100 mol dm-3 Na2S2O3 was required to discharge
the iodine colour.

(a) Calculate the amount in moles of iodine produced.

(7.50 × 104 mol)

(b) Hence, calculate the percentage of O3 in the gaseous mixture.

(3.41%)

(a) Amount of Na2S2O3 = 0.100  (15.0  103) = 1.50  103 mol

Mole ratio of Na2S2O3 : I2 = 2 : 1
Amount of I2 produced = 0.5 × (1.50 × 10−3) = 7.50 × 10−4 mol

(b) Mole ratio of O3 : I2 = 1 : 1

Amount of O3 = 7.50 × 10−4 mol
Volume of O3 = (7.50 × 10−4 ) × 22.7 = 17.025 cm3
Percentage of O3 in the mixture = (17.025 ÷ 500) × 100% = 3.41 %

CHEM~IS~TRY

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