Considering ammonia production with 53% yield, what mass of nitrogen and what mass of hydrogen would be needed to produce the actual yield of 100.0 moles of ammonia?
I know I need 190 moles of NH3 to produce 100 moles of NH3
The formula I'm using right now is 3H2+N2=2NH3

Question

Considering ammonia production with 53% yield, what mass of nitrogen and what mass of hydrogen would be needed to produce the actual yield of 100.0 moles of ammonia?
I know I need 190 moles of NH3 to produce 100 moles of NH3
The formula I'm using right now is 3H2+N2=2NH3

jfraser · Answer

you've done the percent correctly, now you need to turn those moles of NH3 into moles of N2 and moles of H2. Use the molar ratio that the balanced reaction gives you.

2 moles of NH3 are made by 1 mole of N2, so:
$$190mol NH_3 * (\frac{1mol N_2}{2mol NH3}) = 95mol N_2 \space needed$$do the same thing, but with the H2 ratio, and then turn those moles of each reactant into mass using the molar masses of N2 and H2