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%
