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    The answer is 21.2% but i don't know how to get it?

    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?
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    To work out the percentage yield you have to do the (actual amount in moles divided by the theoretical amount in moles) * 100. To work out the actual amount in moles of NH3 we need to do mass / mr, so it is 1.8/17 which gives us 9/85mol. To find the theoretical amount in moles we need to find the moles of N2 and then equate this to 2NH3. So, moles of N2 = 7/28 which gives us 0.25mol and as it is a 1:2 ratio amount of moles of NH3 is 0.5mol. Now we put these values into our equation: ((9/85)/0.5)*100 which gives us 21.1764.....% which is 21.2% to 3 significant figures.
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    (Original post by hamzahussain97)
    To work out the percentage yield you have to do the (actual amount in moles divided by the theoretical amount in moles) * 100. To work out the actual amount in moles of NH3 we need to do mass / mr, so it is 1.8/17 which gives us 9/85mol. To find the theoretical amount in moles we need to find the moles of N2 and then equate this to 2NH3. So, moles of N2 = 7/28 which gives us 0.25mol and as it is a 1:2 ratio amount of moles of NH3 is 0.5mol. Now we put these values into our equation: ((9/85)/0.5)*100 which gives us 21.1764.....% which is 21.2% to 3 significant figures.
    Thank you so much for your help!
 
 
 
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