1.66 g H2 is allowed to react with 10.3 g N2, producing 2.59 g NH3

Question

Part A) What is the theoretical yield in grams for this reaction under the given conditions?

Part B) What is the percent yield for this reaction under the given conditions?

Answer 1

Write a correctly balanced equation:

3H₂ + N₂ ==> 2NH₃

Find the limiting reactant:

moles H₂ = 1.66 g x 1 mole/2 g = 0.83 moles H₂

moles N₂ = 10.3 g x 1 mole/28 g = 0.368 moles N₂

Because the mole ratio of H₂ : N₂ is 3:1, the H₂ is limiting.

Theoretical yield:

0.83 moles H2 x 2 moles NH3/3 moles H2 x 17 g/mole NH3 = 9.41 g NH3

Percent yield:

2.59 g/9.41 g (x100%) = 27.5%