# How much ammonia (NH3) can be obtained when 3.0g H2 reacts with 100g N2?

Dec 1, 2015

Approx. $8 \cdot g$.

#### Explanation:

We need (i), a reaction scheme:

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \rightarrow N {H}_{3} \left(g\right)$

And (ii), molar quantities of each reagent:

${H}_{2}$, $\frac{3.0 \cdot g}{2 \cdot g \cdot m o {l}^{-} 1}$ $=$ $\frac{3}{2} \cdot m o l$ ${H}_{2}$;

${N}_{2}$, $\frac{100.0 \cdot g}{28 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.3 \cdot m o l$ ${N}_{2}$;

Clearly, dihydrogen is in deficiency, and dinitrogen is in vast excess. So at most we can make $\frac{1}{3}$ $\times$ $\frac{3}{2} \cdot m o l$ $=$ $\frac{1}{2}$ $m o l$ ammonia, $\approx$ $8 \cdot g$.

Vast quantities of ammonia are synthesized industrially. Why is this an important reaction?