# Na_3N decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced, how many moles of Na_3N was used (22.4 L = 1 mole of any gas) ?

Jul 29, 2017

$1.22$ mol $N {a}_{3} N$

#### Explanation:

Given: $N {a}_{3} N$ decomposes to form sodium (Na) and Nitrogen (N_2) at STP (standard temperature and pressure). At STP $22.4 L = 1$ mole of gas

First you need to write a skeleton equation and then balance it. You need to realize that nitrogen is a diatomic element ${N}_{2}$.

skeleton equation: $\text{ } N {a}_{3} N \to N a + {N}_{2}$

balanced: $\text{ } 2 N {a}_{3} N \to 6 N a + {N}_{2}$

The balanced equation gives us the mole ratio of sodium nitrate to nitrogen produced: $\frac{2}{1}$.

$13.7 \cancel{L {N}_{2}} \times \frac{1 \cancel{m o l {N}_{2}}}{22.4 \cancel{L {N}_{2}}} \times \frac{2 m o l N {a}_{3} N}{1 \cancel{m o l {N}_{2}}} = 1.22 \text{ mol } N {a}_{3} N$