Question

If 23.6g of nitrogen are mixed with 17.3g of hydrogen, how many grams of ammonia may form?

Answer 1

Balanced chemical reaction:

N2 + 3H2 --------> 2NH3

Thus 1 mole of N2 reacts with 3 moles of H2 to give 2 moles of NH3.

First, we need to decide which is limiting reagent.

23.6 g of N2 = 23.6 g / 28.01 g/mol = 0.8426 moles

17.3 g of H2 = 17.3 g / 2.01 g/mol = 8.6 mol

Thus N2 is the limiting reagent and H2 is in excess.

So, 0.8426 moles of N2 will produce = 0.8426 * 2 = 1.6852 moles of NH3.

1.6852 moles of NH3 = 1.6852 mol *17.03 g/mol = 28.70 g of NH3

Thus 28.07 g of ammonia may form in the given reaction.