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Chemistry question need help

O tonne of ammonia is reacted with carbon dioxide to prepare the fertiliser urea, NHCONH.
2NH3(g) + CO2(g) NH2CONH2(s) + H20(1)

1.35 tonnes of urea are formed

Calculate percentage yield of urea

Show all working out.

In the mark scheme they do theoretical yield / actual yield x100 which is the wrong formula so I’m confused
Reply 1
Must be a mistake in the MS - it should be actual yield/ theoretical yield x 100

I assume you you meant to say 1 tonne of ammonia not 0? So what you do is find the moles of ammonia which is mass (which is equal to 1×10^6 g) divided by the MR which is 17. This gives you approximately 58,823 moles. But because the mole ratio is 2:1 between ammonia and urea, 58823 / 2 is the theoretical yield (29411 moles) of urea. The actual yield of urea is 1.35×10^6 g divided by the Mr (60) which gives 22,500 moles. You then use the equation, actual yield over theoretical yield, which will be 22,500 divided by approximately 29,411, then multiply by 100 to get a percentage. the percentage yield of urea is then 76.5%.
This is how i would do it, I hope that helps, what is the mark scheme answer?

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