# What mass of ammonia can be produced by the reaction of "2.16 mol N"_2" and excess "H"_2"?

Sep 26, 2016

The reaction will produce $\text{73.6 g NH"_3}$.

#### Explanation:

Balanced Equation

$\text{3H"_2"+N"_2}$$\rightarrow$$\text{2NH"_3}$

Molar mass $\text{NH"_3}$: $\text{17.03052 g/mol}$ https://pubchem.ncbi.nlm.nih.gov/compound/222#section=Top

Mole ratios between $\text{NH"_3}$ and $\text{N"_2}$.

$\left(1 {\text{mol N"_2)/(2"mol NH}}_{3}\right)$ and $\left(2 \text{mol NH"_3)/(1"mol N"_2}\right)$

Multiply the given moles of nitrogen gas by the mole ratio that has $\text{NH"_3}$ in the numerator. This gives the moles of $\text{NH"_3}$ produced in this reaction. Then multiply by the molar mass of $\text{NH"_3}$ to get the mass of ${\text{NH}}_{3}$ produced in this reaction.

$2.16 \cancel{\text{mol N"_2xx(2cancel"mol NH"_3)/(1cancel"mol N"_2)xx(17.03052"g NH"_3)/(1cancel"mol NH"_3)="73.6 g NH"_3}}$