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# GCSE Percentage Yield question watch

1. Stuck on this question, could someone explain the steps how to work this out.

Ammonia Gas (NH3) is made by heating the gases nitrogen and hydrogen under pressure in the prescence of an iron catalyst:

N2 + 3H2 ⇌ 2NH3

If 7.0g of nitrogen are reacted with excess hydrogen and 1.8g of ammonia is collected, what is the percentage yield?

The mark scheme says the answer is 21%(21.2%) but I cant get this answer
2. So you have 7g of N2
N2 = 2x14 = RMM of 28

moles of N2 is mass/RMM = 7/28 = 0.25 moles

Convert 2NH3 into grams

2x(14+3) = RMM of 34

Since there was 0.25 moles of nitrogen gas present, multiply this by 0.25

0.25 x 34 = 8.5g of ammonia (theoretical yield)

percentage yield = actual yield/theoretical yield x100

= 1.8/8.5 x100 = 21.18%
So you have 7g of N2
N2 = 2x14 = RMM of 28

moles of N2 is mass/RMM = 7/28 = 0.25 moles

Convert 2NH3 into grams

2x(14+3) = RMM of 34

Since there was 0.25 moles of nitrogen gas present, multiply this by 0.25

0.25 x 34 = 8.5g of ammonia (theoretical yield)

percentage yield = actual yield/theoretical yield x100

= 1.8/8.5 x100 = 21.18%
Thanks! understand it now
So you have 7g of N2
N2 = 2x14 = RMM of 28

moles of N2 is mass/RMM = 7/28 = 0.25 moles

Convert 2NH3 into grams

2x(14+3) = RMM of 34

Since there was 0.25 moles of nitrogen gas present, multiply this by 0.25

0.25 x 34 = 8.5g of ammonia (theoretical yield)

percentage yield = actual yield/theoretical yield x100

= 1.8/8.5 x100 = 21.18%
I was so stuck on this question as well so thank you!!
5. Hi! I moved this to the Maths forum
6. (Original post by Fox Corner)
Hi! I moved this to the Maths forum
this is chemistry...
7. (Original post by nega166)
this is chemistry...
woops - moved to Chemistry now. Sorry!
8. Isnt the mole ratio different in this case, one mole of N2, giving you 2moles of NH3

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