# Question f05aa

Sep 16, 2016

$\text{18.6 g NaN"_3}$ are required to form $\text{12.0 g N"_2}$

#### Explanation:

Balanced Equation

"2NaN"_3("s")$\rightarrow$"2Na(s)"+"3N"_2("g")

We will be doing the following pattern to answer this question.

$\text{mass N"_2}$$\rightarrow$$\text{mol N"_2}$$\rightarrow$$\text{mol NaN"_3}$$\rightarrow$$\text{mass NaN"_3}$

We need the molar ratios between ${\text{NaN}}_{3}$ and ${\text{N}}_{2}$.

From the equation, we can see that there are two mole ratios:

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

We need to determine the molar masses of the nitrogen gas and sodium azide.

Molar Mass $\text{N"_2} :$(2xx"14.0067 g/mol)="28.0134 "g/mol N"_2"#

Molar Mass $\text{NaN"_3:}$$\text{65.009869 g/mol NaN"_3}$
https://www.ncbi.nlm.nih.gov/pccompound?term=NaN3

Next we need to determine how many moles are in $\text{12.0 g N"_2}$ by dividing Its given mass by its molar mass. We will need to do the same for $\text{NaN"_3}$.

Moles $\text{N"_2} :$$12.0 \cancel{\text{g N"_2xx"1 mol N"_2/(28.0134 cancel"g N"_2)="0.4284 mol N"_2}}$

To determine the moles of $\text{NaN"_3}$, we need to multiply the mole ratio from the equation. We will need to use the mole ratio from the equation with $\text{NaN"_2}$ in the numerator.

$0.4284 \cancel{\text{mol N"_2xx(2"mol NaN"_3)/(3cancel"mol N"_2)="0.2856 mol NaN"_3}}$

Now we need to multiply the moles $\text{NaN"_3}$ by its molar mass.

$0.2856 \cancel{\text{mol NaN"_3""xx(65.009869"g NaN"_3)/(1cancel"mol NaN"_3)="18.6 g NaN"_3}}$ (rounded to three significant figures.