# Question 928a6

Sep 14, 2016

The reaction produces 3.60 mol or ${\text{82 dm"^3color(white)(l) "of N}}_{2}$.

#### Explanation:

Step 1. Write the balanced chemical equation.

The balanced equation is

${\text{2NaN"_3 → "2Na" + "3N}}_{2}$

Step 2. Strategy

The problem is to convert grams of ${\text{NaN}}_{3}$ to moles of ${\text{N}}_{2}$ and volume of ${\text{N}}_{2}$.

We can use the flow chart below to help us. The process is:

(a) Use the molar mass to convert the mass of ${\text{NaN}}_{3}$ to moles of ${\text{NaN}}_{3}$.

(b) Use the molar ratio (from the balanced equation) to convert moles of ${\text{NaN}}_{3}$ to moles of ${\text{N}}_{2}$.

(e) Use the Ideal Gas Law to convert moles of ${\text{N}}_{2}$ to volume of ${\text{N}}_{2}$.

In equation form,

${\text{grams of NaN"_3 stackrelcolor(blue)("molar mass"color(white)(ml)) (→) "moles of NaN"_3 stackrelcolor(blue)("molar ratio"color(white)(ml))→ "moles of N"_2 stackrelcolor(blue)("Ideal Gas Law"color(white)(ml))(→) "volume of N}}_{2}$

The Calculations

(a) Moles of ${\text{NaN}}_{3}$

156 color(red)(cancel(color(black)("g NaN"_3))) × ("1 mol NaN"_3)/( 65.01 color(red)(cancel(color(black)("g NaN"_3)))) = "2.40 mol NaN"_3 

(b) Moles of ${\text{N}}_{2}$

2.40color(red)(cancel(color(black)("mol NaN"_3))) × ("3 mol N"_2)/(2 color(red)(cancel(color(black)("mol NaN"_3)))) = "3.60 mol N"_2

(c) Volume of ${\text{N}}_{2}$

The Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this to give

$V = \frac{n R T}{P}$

$n = \text{3.60 mol}$
$R = \text{8.314 Pa·m"^3·"K"^"-1""mol"^"-1}$
$T = \text{(27 + 273.15) K" = "300.15 K}$
P = 1.1 × 10^5color(white)(l) "Pa"

V = (3.60 color(red)(cancel(color(black)("mol"))) × 8.314 color(red)(cancel(color(black)( "Pa")))·"m"^3·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 300.15color(red)(cancel(color(black)( "K"))))/(1.1 × 10^5 color(red)(cancel(color(black)("Pa")))) = "0.082 m"^3 = "82 dm"^3#

The volume of ${\text{N}}_{2}$ produced is ${\text{82 dm}}^{3}$.