Question

Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements: 2NaN3(s)→2Na(s)+3N2(g)

Answer 1

Answer:

$\boxed{\text{(a) 2.85 mol; (b) 21.7 g; (c) 712 g}}$

Explanation:

We will need a balanced chemical equation with molar masses, volumes, and concentrations, so, let's gather all the information in one place.

M_r:     65.01                         28.01

         2NaN₃(s) ⟶ 2Na(s) + 3N₂(g)

n/mol:   1.90

m/g:                                        14.0

(a) Moles of N₂

The molar ratio is 3 mol N₂ = 2 mol NaN₃

$\text{Moles of N}_{2} = \text{1.90 mol NaN _{3} }\times \dfrac{\text{3 mol N}_{2}}{\text{2 mol NaN _{3} }}= \textbf{2.85 mol N}_{2}\\\\\text{The reaction produces \boxed{\textbf{2.85 mol N}_{2}} }$

(b) Mass of NaN₃

(i) Moles of N₂

$\text{Moles of N}_{2} = \text{14.0 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{ 28.01 g N}_{2}} = \text{0.4998 mol N}_{2}$

(ii) Moles of NaN₃

$\text{Moles of NaN}_{3} =\text{0.4998 mol N}_{2} \times \dfrac{\text{2 mol NaN}_{3}}{\text{3 mol N}_{2}} = \text{0.3332 mol NaN}_{3}$

(iii) Mass of NaN₃

$\text{Mass of NaN}_{3} = \text{0.3332 mol NaN}_{3} \times \dfrac{\text{65.01 g NaN}_{3}}{\text{1 mol NaN}_{3}} =\textbf{21.7 g NaN}_{3}\\\\\text{You must use }\boxed{\textbf{21.7 g NaN}_{3}}$

(c)Mass of NaN₃

(i) Volume of N₂

$V = \text{13.0 ft}^{3} \times \dfrac{\text{28.32 L}}{\text{1 ft}^{3}} = \text{368.2 L}$

(ii) Mass of N₂

$\text{Mass of N}_{2} = \text{368.2 L N}_{2} \times \dfrac{\text{1.25 g N}_{2}}{\text{1 L N}_{2}} = \text{460.2 g N}_{2}$

(iii) Moles of N₂

$\text{Moles of N}_{2} = \text{460.2 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{ 28.01 g N}_{2}} = \text{16.43 mol N}_{2}$

(iv) Moles of NaN₃

$\text{Moles of NaN}_{3} =\text{16.43 mol N}_{2} \times \dfrac{\text{2 mol NaN}_{3}}{\text{3 mol N}_{2}} = \text{10.95 mol NaN}_{3}$

(v) Mass of NaN₃

$\text{Mass of NaN}_{3} = \text{10.95 mol NaN}_{3} \times \dfrac{\text{65.01 g NaN}_{3}}{\text{1 mol NaN}_{3}} =\textbf{712 g NaN}_{3}\\\\\text{You must use }\boxed{\textbf{712 g NaN}_{3}}$