# How many grams of H_2 are needed to produce 13.68g of NH_3?

Feb 23, 2016

We need (i) the molar quantity of ammonia, and (ii) a stoichiometric equation that shows ammonia synthesis.

#### Explanation:

$\text{Moles of ammonia}$ $=$ $\frac{13.68 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.803$ $m o l .$

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

From the stoichiometry, we know that there are 1.5 moles dihydrogen required for each mole of ammonia. There we require $1.21$ $m o l$ dihydrogen.

$\text{Mass of dihydrogen} = 1.21 \cdot m o l \times 2.02 \cdot g \cdot m o {l}^{-} 1$ $=$ ?? $g$.