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?

Explanation:

1) Mole ratios:

The coefficients on the balanced chemical equation give the mole ratios:

2) Theoretical yield:

       n = mass in grams / molar mass = 7.81 g / 24.305 g/mol = 0.321 mol

       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

3) Percent yield, %

false

because it does not have the potential to be entirely used up

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