Stoichiometry Problem

Question

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

To solve this problem, we will use three-step stoichiometry twice.

First, let us begin with the 135g of nitrogen gas:

135g N₂ X 1 mol N₂ / 28.02g N₂ X 2 mol NH₃ / 1 mol N₂ X 17.031 g NH₃ / 1 mole = 164g NH₃

Next, perform three-step stoichiometry beginning with 53.1g of hydrogen gas:

53.1g H₂ X 1 mol H₂ / 2.016g H₂ X 2 mol NH₃ / 3 mol H₂ X 17.031g NH₃ / 1 mole = 299 g NH₃

How do you know which of the two answers to choose? Always select the smaller of the two possible answers.