What is the limiting reactant when 3.41 g of nitrogen react with 2.79g of hydrogen to produce ammonia and how many grams of ammonia are produced?

The reaction for the Haber process is

N2 + 3 H2 ⇌ 2 NH3

Amount of N2 = 3.41/28.0 = 0.122mol

Amount of H2 = 2.79/2.0 = 1.40mol

According to the stoichiometry of the reaction, 1 mol of N2 reacts with 3 mol of H2. 0.122mol of N2 will therefore react with only 0.366mol of H2, but there is 1.40mol of H2 available. Thus H2 is in excess and N2 is the limiting reactant.

1mol of N2 reacts to form 2 mol of NH3.

Under the maximum possible yield, 0.122mol of N2 reacts to form 0.244mol of NH3.

Mass of NH3 = 0.244 x 17.0 = 4.15g