Chapter 10, Problem 10.135QA

Interpretation Introduction

To calculate:

a) The volume of $N_2$ produced at ${17}^{0}C$ and $1.00 atm$ if the denitrification process were complete.

b) The volume of $C{O}_2$ produced

c) The density of the gas mixture

Answer to Problem 10.135QA

Solution:

a) $38.4 L N_2$

b) $192 L C{O}_2$

c) $1.74\frac{g}{L}$

Explanation of Solution

1) Concept:

The moles of $N{O}_3^{-}$ can be calculated using its molar mass. The moles of $N_{2}and C{O}_2$ gases can be calculated using the stoichiometry of the balanced reaction.

Using the ideal gas law, the volume of each gas can be calculated.

The total mass of the gases and the total volume of the gas can be used to calculate the density of the gas mixture.

2) Formula:

$PV=nRT$

3) Given:

i) mass of $N{O}_3^{-}=200.0 g$

ii) $T={17}^{0}C=17+273=290 K$

iii) $P=1.00 atm$

iv) Balanced reaction: $2 N{O}_3^{-} \left(aq\right)+5 CO\left(g\right)+2H^{+}\left(aq\right)\to N_2\left(g\right)+H_{2}O \left(l\right)+5 C{O}_2\left(g\right)$

4) Calculations:

Calculating the moles of $N{O}_3^{-}$ and then moles of $N_2$ gas:

From the balanced reaction, it is seen that $2$ moles of $N{O}_3^{-}$ produce $1$ mole of $N_2$ gas.

$200.0 g N{O}_3^{-} \times \frac{1 mol N{O}_3^{-}}{62.004 g N{O}_3^{-} } \times \frac{1 mol N_2}{2 mol N{O}_3^{-}}=1.6128 mol N_2$

a) Calculating the volume of $N_2$ gas:

$PV=nRT$

$V=\frac{nRT}{P}$

$V=\frac{\left( 1.6128 mol\right)\left(0.08206 \frac{L.atm}{K.mol}\right)\left(290 K\right)}{\left(1.00 atm\right)}=38.38 L N_2$

The volume of $N_2$ produced is $38.4 L$

b) Calculating the volume of $C{O}_2$ gas produced:

From the balanced reaction, it is seen that $2$ moles of $N{O}_3^{-}$ produce $5$ mole of ${CO}_2$ gas.

$200.0 g N{O}_3^{-} \times \frac{1 mol N{O}_3^{-}}{62.004 g N{O}_3^{-} } \times \frac{5 mol {CO}_2}{2 mol N{O}_3^{-}}=8.064 mol {CO}_2$

$PV=nRT$

$V=\frac{nRT}{P}$

$V=\frac{\left( 8.064 mol\right)\left(0.08206 \frac{L.atm}{K.mol}\right)\left(290 K\right)}{\left(1.00 atm\right)}=191.9 L {CO}_2$

The volume of ${CO}_2$ produced is $192 L$

c) Calculating the density of the gas mixture:

At ${17}^{0}C and 1.00 atm$, moles of $N_2 and C{O}_2$ gases are known.

Mass of $N_2$ gas =

$1.6128 mol N_2\times \frac{28.014 g N_2}{1 mol N_2}=45.18 g N_2$

Mass of ${CO}_2$ gas =

$8.064 mol {CO}_2\times \frac{44.009 g {CO}_2}{1 mol {CO}_2}=354.89 g {CO}_2$

Total mass of gases $=45.18g+354.89 g=400.07 g$

Total volume of gases $38.4 L+192 L=230.4 L$

Density of gas mixture $= \frac{400.07g }{230.4 L}=1.736\frac{g}{L}$

Thus, the density of the gas mixture is $1.74\frac{g}{L}$

Conclusion:

The moles of the gases are calculated using the stoichiometry of the balanced reaction. The volumes of the gases are calculated using the ideal gas equation and the density of the gas mixture is calculated.