Answer to Question #116037 in Chemistry for Sara Sandoval

Question

For the following reaction, what is the percentage yield of magnesium nitride if 1.70 g of it are produced when 1.36 g of Mg reacts with 0.695 g of N2? (10 pts)
3Mg + N2 —> Mg3N2

Answer 1

3Mg + N₂ —> Mg₃N₂

1. Calculating limiting reactant:

n(Mg) = m(Mg) / [3 × Mr(Mg)] = 1.36 g / (3 × 24 g/mol) = 0.019 mol

n(N₂) = m(N₂) / Mr(N₂) = 0.695 g / 28 g/mol = 0.025 mol

As n(Mg) < n(N₂), Mg is a limiting reactant.

2. Calculating the theoretical yield of Mg₃N₂:

m(Mg₃N₂) = m(Mg) × Mr(Mg₃N₂) / [3 × Mr(Mg)] = n(Mg) × Mr(Mg₃N₂) = 0.019 mol × 101 g/mol = 1.92 g

3. Calculating the percentage yield:

As the theoretical yield equals 1.92 g and actual yield equals 1.70 g, then:

Percentage yield = actual yield / theoretical yield × 100% = 1.70 g / 1.92 g × 100% = 88.5%

Answer: 88.5%