# Question bf4d3

Apr 9, 2017

41%

#### Explanation:

First, we need to calculate theoretical yield from the balanced equation.

$4 N {H}_{3} + 6 N O \to 5 {N}_{2} + 6 {H}_{2} O$ this shows that we should get 5 moles of ${N}_{2}$ for every 4 moles of $N {H}_{3}$ reacted.
10.0g of $N {H}_{3}$ is $\left(\frac{10.0}{17}\right) = 0.588$ moles. The resulting ${N}_{2}$ should then be $\frac{5}{4} \cdot 0.588 = 0.735$ moles. From this value we can either calculate grams, for a mass/mass % yield, or convert our given grams into moles for a mole/mole (volume/volume) % yield.

Mass basis: $0.735 \text{mol} {N}_{2} \cdot \frac{28 g {N}_{2}}{m o l {N}_{2}} = 20.58$ g ${N}_{2}$ Theoretical (8.50/20.58 * 100) = 41.3% by mass

Mole basis: $\frac{8.50 g {N}_{2}}{\frac{28 g {N}_{2}}{m o l {N}_{2}}} = 0.304$ moles ${N}_{2}$ (0.304/0.735 * 100) = 41.4%# by volume