Question

1.15 g H2 is allowed to react with 10.0 g N2, producing 2.24 g NH3.

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

N2 + 3H2 ---> 2NH3

Part A.

moles of H2 = g/molar mass = 1.15 g/2.016 g/mol = 0.570 mols

moles of N2 = 10.0 g/28 g/mol = 0.357 mols

Limiting reagent

If all of N2 reacts we would need = 3 x 0.357 = 1.071 mols H2

If all of H2 reacts we would need = 0.570/3 = 0.190 mols N2

Since moles of H2 are less then required, H2 is the limiting reagent

Theoretical yield of NH3 = 0.570 mol x 2 x 17.031 g/mol/3 = 6.47 g

Part B.

Percent yield = (experimental yield/theoretical yield) x 100

                      = (2.24/6.47) x 100

                      = 34.62%