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?

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%