Corrected question:
Since the question has some errors, I copy it with the corrections:
If 7.81 g of Mg undergoes the following reaction what is the % yield, if 6.76 g of magnesium nitride is produced?
- 3 Mg(s) + N₂ (g) → Mg₃N₂ (s).
Answer:
Explanation:
1) Mole ratios:
The coefficients on the balanced chemical equation give the mole ratios:
- 3 mol Mg(s) : 2 mol N₂ (g) : Mg₃N₂
2) Theoretical yield:
- Number of moles of Mg that react, n:
n = mass in grams / molar mass = 7.81 g / 24.305 g/mol = 0.321 mol
- Number of moles of magnesium nitride that can be produced
Proportion: 3 mol Mg / 1 mol Mg₃N₂ = 0.321 mol Mg / x
⇒ x = 0.321 mol Mg × 1 mol Mg₃N₂ / 3 mol Mg = 0.107 mol Mg₃N₂
- mass = molar mass × number of moles = 100.9494 g/mol × 0.107 mol = 10.80 g
3) Percent yield, %
- % = (actual yield / theoretical yield) × 100 = (6.76 g / 10.80 g) × 100 = 62.6%
