Chapter 11, Problem 11.68E

Interpretation Introduction

Interpretation:

The volume of air that would be needed to combust $1.00\text{g}$ of methane is to be calculated.

Concept introduction:

The stoichiometry of a chemical species involved in a chemical reaction represents the number of chemical species involved in the chemical reaction. The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product. The stoichiometry of a chemical species is also represented by number of moles. The number of moles of a substance is given as,

$n=\frac{m}{M}$

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Answer to Problem 11.68E

The volume of air that would be needed to combust $1.00\text{g}$ of methane is $13\text{L}$.

Explanation of Solution

The complete combustion reaction of methane is represented as,

${\text{CH}}_4+2{\text{O}}_2\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

The given mass of methane is $1.00\text{g}$.

The molar mass of oxygen is $32.0\text{g}/\text{mol}$

The molar mass of methane is $16.0\text{g}/\text{mol}$.

The number of moles of a substance is given as,

$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Substitute the mass and molar mass of methane in the equation (1).

$\begin{array}{rll}n& =& \frac{1.00\text{g}}{16.0\text{g}/\text{mol}}\\ & =& 0.0625\text{mol}\end{array}$

Therefore, the number of moles of methane present in the reaction mixture is $0.0625\text{mol}$.

One mole of methane reacts with two moles of oxygen gas. Therefore, the relation between the number of moles of methane and oxygen is given as,

$n_{{\text{O}}_{\text{2}}}=2n_{{\text{CH}}_4}$ …(2)

Where,

• $n_{{\text{CH}}_4}$ represents the number of moles of methane.

• $n_{{\text{O}}_{\text{2}}}$ represents the number of moles of oxygen.

Substitute the value of $n_{{\text{CH}}_4}$ in the equation (2).

$\begin{array}{rll}{n}_{{\text{O}}_{\text{2}}}& =& 2\left(0.0625\text{mol}\right)\\ & =& 0.125\text{mol}\end{array}$

The pressure of a substance at STP is $1.00\text{atm}$.

The temperature of a substance at STP is $273\text{K}$.

The ideal gas equation is given as,

$PV=nRT$ …(4)

Where,

• $V$ represents the volume occupied by the ideal gas.

• $P$ represents the pressure of the ideal gas.

• $n$ represents the number of moles of the ideal gas.

• $T$ represents the temperature of the ideal gas.

• $R$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}$.

Rearrange the above equation for the value of $V$.

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

Substitute the value of number of moles of oxygen, $P$, $T$ and $R$ in the equation (4).

$\begin{array}{rll}V& =& \frac{\left(0.125\text{mol}\right)\left(0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\right)\left(273\text{K}\right)}{\left(1.0\text{atm}\right)}\\ & =& 2.8003\text{L}\end{array}$

Therefore, the volume of oxygen required to react with methane is $2.8003\text{L}$.

It is given that air is composed of about $21\%\text{v}/\text{v}$ oxygen $\left({\text{O}}_2\right)$.

The formula for volume percentage is represented as:

$\text{Volume}\%=\frac{V}{V_{\text{t}}}\times 100\%$

Where,

• $V$ represents the volume of the compound present in the mixture.

• $V_{\text{t}}$ represents the total volume of mixture.

Rearrange the above equation for the value of $V_{\text{t}}$.

$V_{\text{t}}=\frac{V}{\text{Volume}\%}\times 100\%$ …(5)

Substitute the value of volume percentage of oxygen and volume of oxygen in the equation (5).

$\begin{array}{rll}{V}_{\text{t}}& =& \frac{2.8003\text{L}}{21\%}\times 100\%\\ & =& 13.33\text{L}\\ & \approx & 13\text{L}\end{array}$

Therefore, the volume of air that would be needed to combust $1.00\text{g}$ of methane is $13\text{L}$.

Conclusion

The volume of air that would be needed to combust $1.00\text{g}$ of methane is $13\text{L}$.