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Chemistry calculation question : The Haber Process watch

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    Been stuck on this question on an online past paper, any help would be appreciated :

    ' the percentage yield of Ammonia is the percentage, by mass, of the Nitrogen and Hydrogen which has been converted to Ammonia. Calculate the mass, in tonnes, of ammonia which can be produced from 90 tonnes of hydrogen when the percentage yield is 50%. '

    The mark scheme says that the answer is 225 tonnes but I am not sure how they got there.
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    Start (as always) with the balanced equation.

    Then work out the maximum mass of ammonia that could could be made from 90 tonnes of ammonia, assuming 100% yield and excess nitrogen. You probably want to work out the number of moles of hydrogen to start with.
    To make this stage more simple you could scale your calculation to deal with 90 grams of hydrogen - just don't forget to turn your final answer back into tonnes!

    Now you have the mass of ammonia for 100% yield, just multiply by the actual yield of 50%.
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    (Original post by TutorsChemistry)
    Start (as always) with the balanced equation.

    Then work out the maximum mass of ammonia that could could be made from 90 tonnes of ammonia, assuming 100% yield and excess nitrogen. You probably want to work out the number of moles of hydrogen to start with.
    To make this stage more simple you could scale your calculation to deal with 90 grams of hydrogen - just don't forget to turn your final answer back into tonnes!

    Now you have the mass of ammonia for 100% yield, just multiply by the actual yield of 50%.
    Thanks for the explanation, I have already tried this method but I must be going wrong somewhere because I can't get the answer of 225 tonnes. I did this:

    the balanced equation is N2 + 3H2 ---> 2NH3
    the Mr for hydrogen is 6 ( I did (2x1) x3 ) and then the mass is 90 tonnes (90000000 grams) so I did mass/Mr to get moles which gave 15000000mol.
    then the ratio between the hydrogen and the ammonia is 3:2 so that means there's 10000000mol of ammonia.
    Mr of ammonia is 34 and 10000000 x 34 = 34000000 this would be the mass if it was 100% yield of ammonia
    divided by 2 to give 50% yield gives 17000000 = 170 tonnes

    I still don't get 225 tonnes could it be a mistake on the mark scheme or have I made a mistake somewhere?
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    (Original post by parkchimchim)
    Thanks for the explanation, I have already tried this method but I must be going wrong somewhere because I can't get the answer of 225 tonnes. I did this:

    the balanced equation is N2 + 3H2 ---> 2NH3
    the Mr for hydrogen is 6 ( I did (2x1) x3 ) and then the mass is 90 tonnes (90000000 grams) so I did mass/Mr to get moles which gave 15000000mol.
    then the ratio between the hydrogen and the ammonia is 3:2 so that means there's 10000000mol of ammonia.
    Mr of ammonia is 34 and 10000000 x 34 = 34000000 this would be the mass if it was 100% yield of ammonia
    divided by 2 to give 50% yield gives 17000000 = 170 tonnes

    I still don't get 225 tonnes could it be a mistake on the mark scheme or have I made a mistake somewhere?

    First observation from your working:

    Mr of hydrogen is 2. There just happens to be 3 moles in the balanced equation.
    Try using 2 and see what happens.
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    (Original post by TutorsChemistry)
    First observation from your working:

    Mr of hydrogen is 2. There just happens to be 3 moles in the balanced equation.
    Try using 2 and see what happens.
    Ah okay, so you don't multiply by the number in front. Thank you, I think I understand now!
 
 
 
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