Limiting Reactants

Question

2 N₂ (g) + 3 H₂ (g)  2 NH₃ (g)

How many grams of NH₃ are produced if 32 grams of N₂ and 5.5 grams of H₂ react?

____________ g NH3

Answer 1

One easy way to identify the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever value is less is the limiting reactant.

2N₂ + 3H₂ ==> 2NH₃ ... balanced equation

For N₂: 32 g N₂ x 1 mol N₂ / 28 g = 1.14 mols N₂ (÷2->0.57)

For H₂: 5.5 g H₂ x 1 mol H₂ / 2 g = 2.75 mols H₂ (÷3->0.92)

Since 0.57 is < 0.92, N₂ is the limiting reactant

Next, we use the MOLES of the limiting reactant (1.14 mols) to find the theoretical yield of NH₃:

1.14 mols N₂ x 2 mols NH₃ / 2 mols N₂ x 17 g NH₃ / mol NH₃ = 19 g NH₃ (2 sig. figs.)