If 7.25 g of Mg undergoes the following reaction, what is the % yield if 6.76 g of magnesium nitride is produced? 3 Mg(s)+N 2 (g) Mg 3 N 2 (s)

Answer 1

This is a simple theoretical yield problem disguised a percent yield problem!

Balanced Reaction:

3Mg + N₂ → Mg₃N₂ (Note: 3 mol of Mg are required to form 1 mol of Mg₃N₂)

Molar Masses:

1. Mg = 24.3 g/mol
2. N₂ = 28.0 g/mol
3. Mg₃N₂= 100.9 g/mol

First, we must calculate the potential theoretical yield of Mg₃N₂, given the starting mass of Mg.

7.25 g Mg (1 mol Mg / 24.3 g Mg) = 0.30 mol Mg (1 mol Mg₃N₂ / 3 mol Mg) = 0.10 mol Mg₃N₂

0.10 mol Mg₃N₂ (100.9 g Mg₃N₂ / 1 mol Mg₃N₂ ) = 10.09 g Mg₃N₂

Second, we can calculate the percent yield, given the actual yield of Mg₃N₂ from the question stem.

PY = (actual yield / theoretical yield) x 100%

= 6.76 g Mg₃N₂ / 10.09 g Mg₃N₂ x 100%

= 67.0 %