# How many grams of ammonia, NH_3, would be formed from the complete reaction of 4.50 moles hydrogen, H_2 in the reaction N_2 + 3H_2 -> 2NH_3?

Aug 1, 2016

Approx. $54 \cdot g \text{ ammonia}$

#### Explanation:

You have almost solved this problem yourself, because you have written a stoichiometrically balanced equation, which establishes equivalent mass:

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

This tells us that $28 \cdot g$ of dinitrogen combines with $6 \cdot g$ dihydrogen to give $34 \cdot g$ of ammonia. I take it that you know how I got these equivalent masses (if no, I am happy to explain).

You have $4.5 \cdot m o l$ dihydrogen, and thus, at most I can form $4.5 \cdot m o l \cdot {H}_{2} \times \left(2 \cdot m o l \text{ ammonia")/(3*mol" dihydrogen}\right)$ $=$ $3 \cdot m o l \cdot \text{ ammonia}$.

$\text{Mass of ammonia}$ $=$ $\text{moles "xx" molar mass}$ $=$ $3 \cdot m o l \times 17.0 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g