# Question #710f4

$4 N {H}_{3} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 4 N O \left(g\right) + 6 {H}_{2} O \left(l\right)$
Nitrous oxide is commonly an oxidation product of ammonia. If this is the reaction you propose, then $3$ $\text{equiv}$ of water result from $2$ $\text{equiv}$ of ammonia.
You react $40 \cdot m o l$ of $N {H}_{3}$, then clearly you get $60 \cdot m o l$ of $O {H}_{2}$. I acknowledge that this might not have been the reaction you had in mind.