This is another chemistry calculation question – which BMAT examiners seem to love. The firsts step shown is completely irrelevant to answering the question so you can just ignore it. There are two ways to try and tackle this question – you could do all the calculations and work out how much carbon monoxide is produced in stage 2 and then use those values to do the working for the stage 3; this would work fine – it’s just a little time consuming. Alternatively, you can see that 1 mole of C is used in stage 2 which produces 2 moles of CO. All these moles are used in stage 3 – you need to look at the ratios of compounds on either side to see the number of moles of CO3produced. As CO and CO2 both have a 3 before them – the ratio is 1:1. This means 2 moles of CO2 are produced, that is 88g of CO2(Mr= 12 + 16 x 2 = 44. Mass = 44 x 2 = 88g).