# For the reaction of ammonia with molecular oxygen, forming nitrogen monoxide and water. How do you calculate the theoretical yield of NO when 17.0g NH_3 reacts with 16.0g O_2?

Oct 5, 2016

Well first you write a stoichiometrically balanced equation:

$2 N {H}_{3} \left(a q\right) + \frac{5}{2} {O}_{2} \left(g\right) \rightarrow 2 N O \left(g\right) + 3 {H}_{2} O \left(l\right)$

#### Explanation:

Given stoichiometric oxygen, ammonia gives an equimolar quantity of $N O$.

$\text{Moles of ammonia}$ $=$ $\frac{17.0 \cdot g}{17.03 \cdot g}$ $\cong$ $1 \cdot m o l$

$\text{Moles of dioxygen}$ $=$ $\frac{16.0 \cdot g}{32.03 \cdot g}$ $\cong$ $0.50 \cdot m o l$

Clearly, dioxygen is in deficiency, and only $\frac{4}{5} \times 0.5 \cdot m o l = 0.40 \cdot m o l$ $N {H}_{3}$ will react.

And thus only $0.40 \cdot m o l$ $N O$ will be produced, i.e. $0.40 \cdot m o l \times 30.01 \cdot g \cdot m o {l}^{-} 1$ $=$ $12.0 \cdot g$ $N O$.