What is the maximum mass of ammonia that can be formed when 43.88 grams of nitrogen gas reacts with 10.62 grams of hydrogen gas

Question

What is the maximum mass of ammonia that can be formed when 43.88 grams of nitrogen gas reacts with 10.62 grams of hydrogen gas according to the following equation? Please round your answer to the nearest 0.01 gram.

N₂ + 3 H₂ = 2 NH₃

(Thank you so much; in advance!)

Answer 1

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

Anytime you are given the amounts of BOTH reactants, you must first find which reactant is limiting. One easy way to do this is to simply divide the MOLES of each reactant by the corresponding coefficient in the balanced equation:

For N₂: 43.88 g N₂ x 1 mol N₂ / 28.01 g = 1.567 mols (÷1->1.57)

For H₂: 10.62 g H₂ x 1 mol H₂ / 2.02 g = 5.257 mols (÷3->1.75)

Since 1.67 is less than 1.75, N₂ is the limiting reactant and the moles of N₂ (1.567) will dictate how much product can be formed.

Maximum yield of NH₃ = 1.567 mols N₂ x 2 mol NH₃ / mol N₂ x 17.03 g NH₃ / mol = 53.37 g NH₃