If 7.81 g of Mg undergoes the following reaction what is the wyield if 6.76 g of magnesium nitride is produced?

3 Mg (s) + Nale) - Mg3N2 (5).

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:

62.6 %

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%