Question

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?

Answer 1

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%