# Ammonia is produced in the reaction 3H_2 + N_2 -> 2NH_3. What mass of NH_3 could be produced if 12.5 H_2 reacts with excess nitrogen?

We assess the reaction...$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right)$
We ASSUME $\left(i\right)$ $12.5 \cdot g$ dihydrogen gas, and $\left(i i\right)$ QUANTITATIVE reaction with stoichiometric dinitrogen. (And the second assumption is a bit unwarranted).
$\text{Moles of dihydrogen} = \frac{12.5 \cdot g}{2.016 \cdot g \cdot m o {l}^{-} 1} = 6.20 \cdot m o l$...and thus $2.07 \cdot m o l$ of dinitrogen are required. We get $4.14 \cdot m o l$ ammonia gas, a mass of $4.14 \cdot m o l \times 17.01 \cdot g \cdot m o {l}^{-} 1 = 70.31 \cdot g$...