1.48 g H2 is allowed to react with 9.97 g N2 , producing 1.81 g NH3 . What is the theoretical yield in grams? What is the percentage yield in grams?

Answer 1

Whenever given the amounts of BOTH reactants, one must first find the limiting reactant. One way to do this is to divided the mols of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

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

For N₂ we have ... 9.97 g N₂ x 1 mol / 28 g = 0.356 mols N₂ (÷1->0.356)

For H₂ we have ... 1.48 g H₂ x 1 mol / 2 g = 0.74 mols H₂ (÷3->0.247)

Since 0.247 is less than 0.356, H₂ is the limiting reactant and the mols of H₂ (0.74 mols) will determine how much product (NH₃)t can be formed.

Theoretical yield:

0.74 mols H₂ x 2 mols NH₃ / 3 mols H₂ x 17 g NH₃ / mol NH3 = 8.39 g NH₃

Percent yield:

1.81 g / 8.39 g (x100%) = 21.6%