Past Paper Questions
June 2008
6. In the manufacture of methanol, the final stage involves the following
reaction:
CO(g) + 2H2(g) → CH3OH(g)
(a)
What is the maximum volume of methanol vapour that can be obtained
from 1000dm3 of hydrogen gas under the conditions of the reaction?
(All gases are measured at the same temperature and pressure.)
volume = ...................................................
7. The industrial production of sodium carbonate is a complicated process
involving many stages.
(a)
The equation for the overall reaction is:
CaCO3(s) + 2NaCl(aq) → Na2CO3(aq) + CaCl2(s)
Calculate the minimum mass of calcium carbonate required to produce 1
million tonnes of sodium carbonate.
(Relative atomic masses: C = 12.0, O = 16.0, Na = 23.0, Ca = 40.0)
(1)
answer = ................................... tonnes
(4)
8. In an experiment to determine the solubility of calcium hydroxide in water,
Jacqui shook excess calcium hydroxide with water in a large, stoppered flask.
She allowed the mixture to stand for 24 hours. She then filtered the mixture to
remove excess solid.
Jacqui took 100 cm3 of the solution for analysis. She titrated 10.0 cm3 portions
of the solution with 0.0500 mol dm-3 hydrochloric acid, HCl, to determine how
much calcium hydroxide had been dissolved. She used methyl orange as the
indicator.
Here are her results.
volume of calcium hydroxide solution used for each titration = 10.0 cm3
volume of hydrochloric acid added trial titration = 9.30 cm3
1st titration = 8.90 cm3
2nd titration = 8.80 cm3
average = 8.85 cm3
(a) Jacqui used methyl orange as the indicator in the titration.
Why is universal indicator not suitable for this titration?
(b) Describe how Jacqui should carry out this titration.
(1)
(4)
(c) The equation for the reaction is:
Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)
Calculate the number of moles of calcium hydroxide, Ca(OH)2, present in the
original 100 cm3 of solution.
You may do the calculation either by following steps (i), (ii) and (iii) or by
using another method in (iii) only.
(i) Calculate the number of moles of hydrochloric acid, HCl, used in the
titration.
(1)
(ii) Use the equation for the reaction to calculate the number of moles of
calcium hydroxide, Ca(OH)2, (in 10.0 cm3 of solution) that reacted with this
amount of hydrochloric acid.
(1)
(iii) Calculate the number of moles of calcium hydroxide, Ca(OH)2, present in
the original 100 cm3 solution.
(1)
.
answer = ...........................................
(d) The relative formula mass of calcium hydroxide is 74.0.
Use your answer to (c)(iii) to calculate the mass of calcium hydroxide
dissolved in the original 100 cm3 solution.
answer = ........................................ g
(1)
(e) The solubility of barium chloride in water cannot be determined by titrating
with an acid.
Describe an alternative method for finding the mass of barium chloride
present in 100 cm3 of saturated solution.
(2)
June 2009
6. (c) Another ester, methyl methanoate, can be made as follows:
HCOOH +
Methanoic
acid
CH3OH → HCOOCH3 +
methanol
methyl
methanoate
H2O
water
(i) Calculate the mass of methyl methanoate, HCOOCH3, that can be made
from 320 kg methanol, CH3OH.
(Relative atomic masses: C = 12, H = 1.0, O = 16)
Answer ..................................... kg
(3)
Extra Questions
1. Over 150 million tonnes of sulphuric acid (H2SO4) are produced
annually, making it one of the world’s most manufactured chemicals.
Sulphuric acid is produced industrially using the Contact process.
The first stage is to react sulphur with oxygen to produce sulphur dioxide gas.
S + O2 → SO2
(a) Calculate the mass of sulphur required to produce 100kg of sulphur
dioxide.
Relative atomic masses: sulphur = 32, oxygen = 16.
(3)
2. Chlorine gas is produced industrially by the electrolysis of brine.
The overall equation for the electrolysis is:
2NaCl + 2H2O → 2NaOH + H2 + Cl2
(a)
In one experiment 11.7g of sodium chloride reacted completely to form
chlorine.
(i) Calculate the number of moles of NaCl present in 11.7g
Relative atomic masses: sodium = 23, chlorine = 35.5
(2)
(ii) How many moles of Cl2 were produced by 11.7g of NaCl?
(1)
(iii) Calculate the volume of Cl2 produced, at room temperature and pressure.
1 mole of gas occupies 24dm3 at room temperature and pressure.
(1)
Answers
June 2008
6. 500 (dm3);
(1)
7. (a) Mr(CaCO3) = 100;
Mr(Na2CO3) = 106;
Mass of CaCO3 = 100; x 1 x 106; = 943 000 (tonnes)
106
[Mr; Mr; scaling; conversion to tonnes with 3-6 sig fig;]
(1)
(1)
(1)
[943,000 = 4 marks
943,396 = 4 marks
0.943… million = 4 marks
1,060,000 = 3 marks
[3-6 sig fig allowed for 4th mark]
8. (a) no sharp change of colour/ range of colours/ OWTTE;
(1)
(b) Any four from:
1. wash burette with acid, wash pipette with calcium hydroxide;
2. acid in burette, calcium hydroxide in pipette;
3. correct use of conical flask;
4. add drops/ small amount of methyl orange (to calcium hydroxide);
5. add acid slowly;
6. until indicator turns orange/red/pink;
7. swirl / stir to mix;
8. use of white tile;
9. repeat process to obtain reliable /concordant results/ results to within 0.2
cm3
[Ignore to give average];
(4)
(c) (i) 8.85 x 0.05 = 0.0004425 moles;
1000
[=4.425 x 10-4, allow 2-4 sig fig]
(ii) 0.000445 = 0.00022125 moles;
(1)
2
[Half of previous answer]
[=2.2125 x 10-4, allow 2-5 sig fig]
(iii) 0.00022125 x 10 = 0.0022125 moles;
(1)
[previous answer x 10]
(1)
[=2.2125 x 10-3, allow 2-5 sig fig]
(iii) Mass of Ca(OH)2 = 0.0022125 x 74 = 0.163725 g;
[previous answer x 74]
[allow 2-6 sig fig]
Evaporate (all 100 cm3 of solution) [if sample, eg 10
cm3 used, must calculate correctly, eg x10];
Weigh residue [must calculate mass of solid, and not
solvent];
(1)
(2)
June 2009
6. Mr (HCOOCH3) = 60
Mr (CH3OH) = 32
(1)
30,000g = 10,000 mol of CH3OH
32
From equation 1 mole reacts with 1 mole:
(1)
Number of moles HCOOCH3 = 10, 000 mol
Mass required = 10, 000mol x 60 = 600,000g = 600kg
(1)
Extra Questions
1 (a) Relative mass of SO2 = 32 + (16 x 2) = 64gmol-1
(1)
Mass of SO2 = 100kg = 100 x 1000 = 100,000g
Number of moles of SO2 = m = 100,000g
= 1562.5mol
Mr 64gmol-1
1 mole of S produces 1 mole of SO2 so:
Number of moles of S = 1562.5mol
(1)
Mr of sulphur = 32 gmol-1
m = Mr x n
m = 32 x 1562.5
m = 50,000g = 50kg
(1)
2. (a) (i) Mass of one mole: 23 + 35.5 = 58.5g
(1)
n=m
M
Number of moles = 11.7g = 0.2 moles
58.5gmol-1
(1)
(2 marks for correct answer, otherwise 1 mark for finding the molar mass of
NaCl)
(ii) Ratio of NaCl: Cl2
2:1
0.2moles = 0.1 moles of Cl2
2
(1)
(iii) 24 x 0.1 = 2.4dm3
(1)