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9 May, 13:02

What percent yield of ammonia is produced from 15.0 kg each of h2 and n2, if 13.7 kg of product are recovered? assume the reaction goes to completion?

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  1. 9 May, 16:53
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    First of all, convert given masses to number of moles:

    H2 = 15 kg / (2 kg / kmol) = 7.5 kmol

    N2 = 15 kg / (28 kg / kmol) = 0.5357 kmol

    NH3 = 13.7 kg / (17 kg / kmol) = 0.8059 kmol

    The balanced chemical reaction is:

    N2 + 3H2 - - > 2NH3

    We can see that N2 is the limiting reactant and for every 1 mole of N2, there are 2 moles of NH3 produced, hence:

    NH3 theoretically produced = 0.5357 kmol * (2 / 1) = 1.0714 kmol

    Therefore the percent yield assuming that the reaction is complete is:

    % yield = (0.8059 kmol / 1.0714 kmol) * 100

    % yield = 75.22%
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