Assume 0.280 mol of N2 and 0.893 mol of H2 are present initially.

Assume 0.280 mol of N2 and 0.893 mol of H2 are present initially.

N2 (g) +3H2 (g) -> 2NH3 (g)

After complete reaction, how many moles of ammonia are produced?

How many moles of H2 remain?

Based on the rx, 1 mol of N2 reacts with 3 moles of H2 or 0.28 mol will react with 0.84 moles of H2, so H2 is in excess by a factor 0.893 - 0.84 = 0.053 moles

Now 1 mol of N2 gives 2 moles of NH3
so 0.28 moles will give 0.28*2 = 0.56 moles of NH3