Chapter 10, Problem 10.150QA

Interpretation Introduction

To calculate:

a) The grams of sodium azide needed to inflate a bag at the given pressure and temperature.

b) How much more sodium azide is needed if the bag must produce the same pressure at ${10}^{0}C$?

Answer to Problem 10.150QA

Solution:

a) $67.6 g$ of sodium azide is needed.

b) $2.40 g$ more $Na{N}_3$ is needed at ${10}^{0}C$.

Explanation of Solution

1) Concept:

The volume of the bag is calculated using its dimensions. Using the ideal gas law, the moles of $N_2$ gas can be calculated. From the stoichiometry of the balanced reaction, the moles of sodium azide can be calculated. The molar mass of sodium azide can be used to convert its moles to grams.

2) Formula:

$PV=nRT$

3) Given:

i) Dimensions of the bag = $40 \times 40\times 20 cm$

ii) Pressure, $P=1.25 atm$

iii) Temperature, $T={20}^{0}C=20+273=293 K$

iv) Temperature, $T={10}^{0}C=10+273=283 K$

4) Calculations:

a) Calculation for the mass of sodium azide needed to inflate the bag at given $20℃$ and 1.25 atm.

Calculating the volume of the bag:

The dimensions of the bag are given. From this, the volume of the bag is

$V=40 cm \times 40 cm \times 20 cm= 32000 c{m}^3$

$32000 c{m}^3 \times \frac{1 mL }{1 c{m}^3} \times \frac{1 L}{{10}^3 mL}=32.0 L$

Calculating the moles of gas $N_2$ gas

$PV=nRT$

$n=\frac{PV}{RT}$

$n= \frac{\left(1.25 atm \right)\left(32.0 L\right)}{\left(0.08206 \frac{L.atm}{K.mol}\right)\left( 293K\right)}=1.6636 mol N_2$

Calculating the moles of sodium azide,$Na{N}_3$:

From the balanced reaction, it is seen that $32$ moles of $N_2$ gas are produced by the reaction of $20$ moles of $Na{N}_3$.

$1.6636 mol N_2 \times \frac{20 mol Na{N}_3}{32 mol N_2}=1.03975 mol Na{N}_3$

Calculating the mass of sodium azide:

$1.03975 mol Na{N}_3 \times \frac{65.011 g Na{N}_3}{1 mol Na{N}_3}=67.59 g Na{N}_3$

$67.6 g Na{N}_3$ of sodium azide is needed to inflate the bag at given $20℃$ and 1.25 atm.

b) Calculation for the mass of sodium azide needed to inflate the bag at the given $10℃$ and 1.25 atm.

The number of moles of $N_2$ would be

$n= \frac{\left(1.25 atm \right)\left(32.0 L\right)}{\left(0.08206 \frac{L.atm}{K.mol}\right)\left( 283K\right)}=1.7224 mol N_2$

$1.7224 mol N_2 \times \frac{20 mol Na{N}_3}{32 mol N_2} \times \frac{65.011 g Na{N}_3}{1 mol Na{N}_3}=69.98 g Na{N}_3$

$70.0 g Na{N}_3$ of sodium azide is needed to inflate the bag at given $20℃$ and 1.25 atm.

The additional mass of $Na{N}_3$ required to produce the same pressure at ${10}^{0}C$ is

$70.0 g-67.6 g= 2.40 g$

Thus, $2.40g$ of additional $Na{N}_3$ is required.

Conclusion:

The ideal gas equation is used to calculate the moles of the $N_2$ gas produced. The stoichiometry of the balanced reaction is used to calculate the amount of sodium azide required.