# What is the percent yield for the reaction between 15.8 g of NH_3 and excess oxygen to produce 21.8 g of NO gas and water?

Apr 23, 2016

$2 N {H}_{3} \left(g\right) + \frac{5}{2} {O}_{2} \left(g\right) \rightarrow 2 N O \left(g\right) + 3 {H}_{2} O \left(l\right)$. The yield is approx. 70%.

#### Explanation:

I assume that $15.8 \cdot g$ of ammonia reacts with ${O}_{2}$ to give $21.8 \cdot g$ $N O$, and water. Ammonia is the limiting reagent.

$\text{Moles of ammonia}$ $=$ $\frac{15.8 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.928 \cdot m o l$

$\text{Moles of } N O$ $=$ $\frac{21.8 \cdot g}{30.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.726 \cdot m o l$

Since the ratio between $N {H}_{3}$ and $N O$ is $1 : 1$, the yield is $\frac{0.726 \cdot m o l}{0.928 \cdot m o l}$ xx100% $=$ ??%