When 56.6 g of calcium and 30.5 g of nitrogen gas undergo a reaction that has a 93.0% yield, what mass (g) of calcium nitride is formed?

Calcium (Ca) reacts with Nitrogen gas (N₂) to form Calcium nitride (Ca₃N₂).

• Molar mass of Ca = $40.08 \text{ g/mol}$
• Molar mass of N₂ = $2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol}$
• Molar mass of Ca₃N₂ = $(3 \times 40.08 \text{ g/mol}) + (2 \times 14.01 \text{ g/mol}) = 120.24 \text{ g/mol} + 28.02 \text{ g/mol} = 148.26 \text{ g/mol}$

• Moles of Ca = $\frac{56.6 \text{ g}}{40.08 \text{ g/mol}} = 1.4122 \text{ mol Ca}$
• Moles of N₂ = $\frac{30.5 \text{ g}}{28.02 \text{ g/mol}} = 1.0885 \text{ mol N}_2$

Using the stoichiometric ratio from the balanced equation ($3 \text{ mol Ca} : 1 \text{ mol N}_2$):

• To react all $1.4122 \text{ mol Ca}$, we need:

Since we have $1.0885 \text{ mol N}_2$, which is more than $0.4707 \text{ mol N}_2$, nitrogen is in excess. Therefore, Calcium (Ca) is the limiting reactant.

The theoretical yield is based on the limiting reactant, Ca. From the balanced equation, $3 \text{ mol Ca}$ produces $1 \text{ mol Ca}_3\text{N}_2$.

• Moles of Ca₃N₂ produced = $1.4122 \text{ mol Ca} \times \frac{1 \text{ mol Ca}_3\text{N}_2}{3 \text{ mol Ca}} = 0.4707 \text{ mol Ca}_3\text{N}_2$
• Mass of Ca₃N₂ (theoretical) = $0.4707 \text{ mol} \times 148.26 \text{ g/mol} = 69.79 \text{ g Ca}_3\text{N}_2$

The percent yield is $93.0\%$.

Rounding to three significant figures, the actual mass of calcium nitride formed is $64.9 \text{ g}$.