# How many grams of H_2 must react, if the reaction needs to produced 63.5g of NH_3?

Feb 6, 2017

We need approx. $12 \cdot g$ of $\text{dihydrogen gas.........}$

#### Explanation:

As a preliminary we require a stoichiometrically balanced equation:

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s N {H}_{3} \left(g\right)$

Is this balanced?

$\text{Moles of ammonia} = \frac{63.5 \cdot g}{17.01 \cdot g \cdot m o {l}^{-} 1} = 3.73 \cdot m o l .$

Given the stoichiometric equation, clearly, we need $\frac{3}{2}$ $\text{equiv}$ of $\text{dihydrogen gas:}$

$\text{i.e.}$ $\frac{3}{2} \times 3.73 \cdot m o l \times 2.02 \cdot g \cdot m o {l}^{-} 1 = 11.3 \cdot g$

How much dinitrogen do we need?