# In the equation N_2(g) + 3H_2(g) -> 2NH_3(g), 8.5 grams of hydrogen and 15.8 grams of nitrogen react to form ammonia. How many grams of ammonia are produced?

Dec 31, 2015

#### Explanation:

Moles of ${N}_{2}$ reactant $=$ $\frac{15.8 \cdot g}{28.02 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.564$ $m o l$.

Moles of ${H}_{2}$ reactant $=$ $\frac{8.5 \cdot g}{2.02 \cdot g \cdot m o {l}^{-} 1}$ $=$ $4.21$ $m o l$.

Dihydrogen is the reagent in excess (clearly!). Given that dinitrogen is the limiting reagent, at most I can form $1.13$ mol ammonia gas (i.e. 2 equiv with respect to dinitrogen).

If I have formed $1.13$ $m o l$ ammonia, this is $1.13$ $\cancel{{\text{mol")xx17.0*g*cancel("mol}}^{-} 1}$ $=$ ??g

You might already know that this is probably the most important reaction on the planet. An industrialist would have an orgasm if he could achieve such yields.