# How many moles of N_2 are formed by the decomposition of 2.5 moles of NaN_3?

Feb 16, 2016

$3.75 m o l {N}_{2}$

#### Explanation:

The decomposition reaction of sodium azide $N a {N}_{3}$ is:

$2 N a {N}_{3} \left(s\right) \to 2 N a \left(s\right) + 3 {N}_{2} \left(g\right)$

Using dimensional analysis, we can find the number of mole of ${N}_{2}$ gas formed:

?molN_2(g)=2.5cancel(mol NaN_3)xx(3molN_2)/(2cancel(molNaN_3))=3.75molN_2

Feb 16, 2016

$5.8 m o l s$

#### Explanation:

first multiply the mols with the mr $65 \cdot 2.5 = 162.5$
then use this to get the mols of ${N}_{2}$
$m o l s = \frac{162.5}{28}$
$m o l s = 5.8$