This is based on the June 2013 exam paper, which students from the 2012J presentation will have and others will be able to buy through OUSA. For copyright reasons I can't reproduce the question in full, so this post will make more sense with the paper.
Question 3
Knowing that this has been a struggle for some, but is really something that should count as a “banker” because all the information except for the final two marks is in the question, I thought it was worth going through in some detail.
Question 3 is based on a sample of coal whose constituents and their proportions are described in the table presented, which also gives the relative atomic mass (weight) for each.
Part a) however doesn't refer directly to the coal, but rather asks for the equation for the complete combustion of carbon in air.
This is simply
C + O2 → CO2
Some minor points to note:
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Complete combustion. This means the partial combustion equation (2C + O2 → 2CO) wouldn't gain any marks.
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Carbon: you don't need to address any other components of the coal.
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Air: although air is a mixture of nitrogen (N2), oxygen (O2) and other gases oxygen is the only significant reactant in the combustion of carbon, so you can ignore the others.
Part b) now asks for the amount of carbon dioxide produced by the combustion of 1 kg of the coal.
The table tells us the coal is 88% carbon, so the amount of carbon involved in the combustion is 88% of 1 kg, or 0.88kg. Once again there is no need to worry about any other components, because the question is about the amount of CO2. Nor is there any need to worry about carbon left unburned, as the question says you can assume zero ash.
CO2 is obviously going to be heavier than the original carbon, and this is where the relative atomic masses come in. 1 “unit” of carbon has an atomic mass of 12. 1 “unit” of oxygen has an atomic mass of 16. So 1 “unit” of CO2 has a mass of 1 carbon + 2 oxygen, or 44.
So the CO2 is 44/12 times as heavy as the carbon we started with.
The mass of CO2 produced therefore is given by
44/12 x 0.88 = 3.23 kg or a little over 3.2 kg as specified in the question.
Part c) is similar to part a), but this time with sulphur rather than carbon:
S + O2 → SO2
Part d) is similar to part b), but again using the figures for sulphur which makes up 1% of coal. The figures are simpler this time: there 0.01 kg or 10g sulphur; sulphur has a relative atomic mass of 32 so the addition of two oxygens at mass 16 per unit gives a total of 64, so the SO2 is simple 64/32 = 2 x the weight of the sulphur we started with or 2 x 10 = 20g.
Part e) is where it gets a little – but not much – more complicated. Key point is that that we need to make some conversions, to ensure we are always using the same units of energy. Remember especially that kWh (unit of energy) is different from a kW (unit of power).
The house requires 10000 kWh per year. First thing, since the heat value of the coal is expressed in GJ per tonne, is to convert this to GJ: the conversion rate is given in the question
10000 kWh = 10000 x 3.6 MJ = 36000 MJ or 36 GJ
The heat energy in the coal is 34 GJ / tonne, so it looks like we need just over a tonne. But the house has open grate fireplaces with a thermal efficiency of 25%, meaning we lose ¾ of the heat, so in fact we will need just over 4 tonnes, or to be more precise
(36 x 4)/34 = 4.2 t to 2 s.f.
Part f) we use this figure and the figures from parts b) and d):
From part b) 1 kg coal produces 3.2 kg of CO2, meaning that 1 t of coal produces 3.2 t of CO2. Similarly 1 kg coal produces 20 g of SO2, meaning that 1t coal produces 20 kg. So this is a simple multiplication:
3.2 x 4.2 = 13.44 t CO2
20 x 4.2 = 84 kg SO2
Finally part g), and there are many possible (inter-related) reasons: rise of central heating (both more efficient and more likely to use gas or oil than coal), discovery of natural gas, switch of railways from coal to diesel and electric traction (though this was almost complete by 1970). I haven't seen the sample answer, so I am not 100% sure what was acceptable.