Stoichiometry Problem

Question

1. What is the limiting reagent when 135 g of Nitrogen reacts with 53.1g of hydrogen in the production of Ammonia? What is the maximum amount (in grams) of Ammonia that can be produced?

                                          N₂ + 3 H₂ → 2NH₃

Answer 1

The simplest way to find the limiting reactant in this type of problem is as follows:

Take moles of each reactant and divide that by the coefficient in the balanced equation. Whichever value is the least will be the limiting reactant.

In this problem, we proceed as follows:

For N₂: 135 g N₂ x 1 mol N₂ / 28 g = 4.82 moles N₂ (÷ 1 -> 4.82)

For H₂: 53.1 g H₂ x 1 mol H₂ / 2 g = 26.6 moles H₂ (÷ 3 -> 8.85)

Since 4.82 is less than 8.85, N2 is the limiting reactant.

The maximum amount of ammonia (NH3) that can be formed is referred to as the theoretical yield, and is dependent only on the amount (moles) of the limiting reactant. We calculate this as follows:

4.82 moles N₂ x 2 mols NH₃ / 1 mol N₂ x 17 g NH₃ / mol NH₃ = 164 g NH₃ = max. grms that can be formed