# How many grams of dihydrogen gas are required to produce a 10.53*g mass of ammonia?

Sep 11, 2016

Almost $2 \cdot g$ of dihydrogen gas are required.

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

#### Explanation:

We follow the given stoichiometric equation, which explicitly states that $14 \cdot g$ dinitrogen gas react with $3.0 \cdot g$ dihydrogen to give $17 \cdot g$ ammonia.

$\text{Moles of ammonia}$ $=$ $\frac{10.53 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.618 \cdot m o l$

And thus $\frac{3}{2} \times 0.618 \cdot m o l \times 2.0159 \cdot g \cdot m o {l}^{-} 1 \text{ dihydrogen}$ are required.