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

the 190 moles you found it the theoretical yield. Use the balanced reaction to convert that to moles of nitrogen and moles of hydrogen $$190mol NH{_3} * (\frac{1mol N{_2}}{2mol NH{_3}})$$ and $$190mol NH{_3} * (\frac{3mol H{_2}}{2mol NH{_3}})$$ This gets you the moles of raw materials that should get you 190 moles of ammonia, but will only get you 100moles due to the 53% yield.