Chapter 13, Problem 97A

If 2.33 L of propane at 24°C and 67.2 kPa is completelyburned in excess oxygen, how many moles of carbondioxide will be produced?

Interpretation Introduction

Interpretation:

The number of moles of carbon dioxide produced by $2.33L$ propane gas is to be calculated.

Concept introduction:

It involves stoichiometry concepts and Avogadro’s law.

Answer to Problem 97A

The moles of carbon dioxide produced is $\text{0}\text{.190}\text{mol}$

Explanation of Solution

The given equation is:

$C_3{H}_8+5O_2\to 4H_{2}O+3C{O}_2$

Equation shows $1L$ of propane gas forms $3LC{O}_2$

Therefore $2.33Lpropane$ will be form carbon dioxide:

$\begin{array}{c}2.33Lpropane=2.33Lpropane\times \frac{3LC{O}_2}{1Lpropane}\\ =6.99LC{O}_2\end{array}$

The number of moles is calculated using ideal gas equation.

According to this law,

$\mathrm{PV}=nRT$

Or

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

Given data is:

$\mathrm{V}=6.99L$

$\text{P=67}\text{.2}\text{kPa}$

$\text{R=8}\text{.314}{\text{LkPaK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

$\text{T=24}\text{°C}$

To convert temperature into Kelvin,

$\begin{array}{c}\mathrm{T}\left(K\right)=T\left(°C\right)+273\\ =24+273\\ =297K\end{array}$

Now substituting the values into formula,

$\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{67.2\times 6.99}{8.314\times 297}\\ =0.19mol\end{array}$

Conclusion

$0.19mol$ carbon dioxide is produced.