Chemistry, 05.09.2019 22:30 emilyturchon

Automotive air bags inflate when sodium azide, nan3, rapidly decomposes to its component elements: 2nan3(s)→2na(s)+3n2(g)
(a) how many moles of n2 are produced by the decomposition of 1.90 mol of nan3?
(b) how many grams of nan3 are required to form 14.0 g of nitrogen gas?
(c) how many grams of nan3 are required to produce 13.0 ft3 of nitrogen gas if the gas has a density of 1.25 g/l?

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Answer from: biju10775

The moles of nitrogen gas produced is 2.85 moles.

21.6 grams of sodium azide will be required.

709 grams of sodium azide will be required.

Given:

An automotive airbag inflates when sodium azide decomposes.

The decomposition reaction:

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

To find:

(a) The moles of N_2 produced by the decomposition of 1.90 mol of NaN_3 .

(b) Grams of NaN_3 are required to form 14.0 grams of nitrogen gas

Solution:

(a)

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

Moles of sodium azide = 1.90 mol

According to reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas, then 1.90 moles of sodium azide will give:

$=\frac{3}{2}\times 1.90 mol=\text{2.85 mol of }N_2$

The moles of nitrogen gas produced is 2.85 moles.

(b)

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

Mass of nitrogen gas = 14.0 g

Moles of nitrogen gas = $\frac{14.0 g}{28.0134 g/mol}=0.500 mol$

According to the reaction, 3 moles of nitrogen gas are obtained from 2 moles of sodium azide, then 0.500 moles of nitrogen gas will be obtained from:

$=\frac{2}{3}\times 0.500 mol=0.333\text{ mol of }NaN_3$

Mass of 0.333 moles of sodium azide:

$=0.333 mol\times 65.0099 g/mol=21.6 g$

21.6 grams of sodium azide will be required.

(c)

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

The volume of nitrogen gas = V = 13.0 ft^3

$1 ft^3=28.317 L\\V=13.0ft^3=13.0\times 28.317 L=368L$

The density of the nitrogen gas = d = 1.25 g/L

The mass of nitrogen gas = m

$d=\frac{m}{v}\\1.25 g/L=\frac{m}{368 L}\\m=1.25 g/L\times 368 =460g$

Mass of nitrogen gas = 460 g

Moles of nitrogen gas = $\frac{460g}{28.0134 g/mol}=16.4mol$

According to the reaction, 3 moles of nitrogen gas are obtained from 2 moles of sodium azide, then 16.4 moles of nitrogen gas will be obtained from:

$=\frac{2}{3}\times 16.4mol=10.9\text{ mol of }NaN_3$

Mass of 10.9 moles of sodium azide:

$=10.9mol\times 65.0099 g/mol=708.60g\approx 709g$

709 grams of sodium azide will be required.

Learn more about the unitary method here:

link

Answer from: carlo123

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

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

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