Question

How many moles of nitrogen would react with excess hydrogen to produce 520 mL of ammonia?

0.0116mol
0.012mol
0.0232mol
0.024mol

Answer 1

N₂ + 3H₂ = 2NH₃

n(N₂)=V(NH₃)/{2Vm}

V(NH₃)=520 mL=0.520L
Vm=22.4 L/mol

n(N₂)=0.520/{2*22.4}=0.0116 mol