Question

Sodium azide, the explosive chemical used in automobile airbags is made by the following reaction:
$NaN{O}_3+3NaN{H}_2⟶Na{N}_3+3NaOH+N{H}_3$
If you combine $15.0gNaN{O}_3$ with $15.0gNaN{H}_2$, what mass of $Na{N}_3$ is produced? (Round to the nearest integer)

Answer 1

The balanced chemical equation for the formation of sodium azide is as given below:
$NaN{O}_3+3NaN{H}_2⟶Na{N}_3+3NaOH+N{H}_3$
$15.0$ g sodium nitrate corresponds to $\frac{15.0}{85}=0.1765moles$.
$15.0$ g of sodamide corresponds to $\frac{15.0}{39}=0.385moles$.
Hence, sodamide is the limiting reagent.
$0.385$ moles of sodamide will produce $\frac{0.385}{3}=0.128moles$ of sodium azide.
The mass of sodium azide produced will be $0.128\times 65=8.3g$.
Hence, $8$ g of $Na{N}_3$ is produced.