# When 56.6 g of calcium is reacted with nitrogen gas, 32.4 g of calcium nitride is produced. What is the percent yield of calcium nitride in the reaction 3Ca(s) + N_2(g) -> Ca_3N_2(s)?

Aug 15, 2016

$\text{% yield}$ $<$ 50%

#### Explanation:

$3 C a \left(s\right) + {N}_{2} \left(g\right) \rightarrow C {a}_{3} {N}_{2} \left(s\right)$

$\text{Moles of calcium metal}$ $=$ $\frac{56.6 \cdot g}{40.08 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.41 \cdot m o l$.

$\text{Moles of calcium nitride}$ $=$ $\frac{32.4 \cdot g}{148.25 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.219 \cdot m o l$.

Given the equation, $0.470 \cdot m o l$ calcium nitride would be obtained if the yield were quantitative.

$\text{Yield}$ $=$ (0.219*mol)/(0.470*mol)xx100% $=$ ??