Chapter 13, Problem 88A

Interpretation Introduction

Interpretation: The mass of water vapor produced by $3\text{L}$ propane gas needs to be determined.

Concept introduction: It involves stoichiometry of the reaction and Avogadro’s law.

Also, according to the ideal gas law:

  $\text{P}\text{V}=\text{n}\text{R}\text{T}$

Or

  $\text{n}=\frac{\text{P}\text{V}}{\text{R}\text{T}}$

Here, n is the number of moles, P is pressure, V is volume, R is Universal gas constant and T is temperature.

Answer to Problem 88A

The mass of water vapor produced is $\text{4}\text{.176 g}$ .

Explanation of Solution

The given equation is:

  ${\text{C}}_3{\text{H}}_8+5{\text{O}}_2\to 4{\text{H}}_2\text{O}+3\text{C}{\text{O}}_2$

Equation shows $1\text{L}$ of propane gas forms $4\text{L}{\text{H}}_2\text{O}$

Therefore $3\text{L}\text{p}\text{r}\text{o}\text{p}\text{a}\text{n}\text{e}$ will be form water vapor:

  $\begin{array}{c}3\text{L}\text{p}\text{r}\text{o}\text{p}\text{a}\text{n}\text{e}=3\text{L}\text{p}\text{r}\text{o}\text{p}\text{a}\text{n}\text{e}\times \frac{4\text{L}{\text{H}}_2\text{O}}{1\text{L}\text{p}\text{r}\text{o}\text{p}\text{a}\text{n}\text{e}}\end{array}$

       $\begin{array}{c}=12\text{L}{\text{H}}_2\text{O}\end{array}$

Now the mass of water vapor can be calculated by the number of moles which is calculated by using the ideal gas equation.

According to this law,

  $\text{P}\text{V}=\text{n}\text{R}\text{T}$

Or

  $\text{n}=\frac{\text{P}\text{V}}{\text{R}\text{T}}$

Given data is:

  $\text{V}=12\text{L}$

  $\text{P}=0.990\text{a}\text{t}\text{m}$

  $\text{R=}0.0821\text{L}\text{a}\text{t}\text{m}{\text{K}}^{-1}\text{m}\text{o}{\text{l}}^{-1}$

  $\text{T=350}\text{°C}$

To convert temperature into Kelvin,

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

   $\begin{array}{c}=350+273\end{array}$

   $\begin{array}{c}=623\text{K}\end{array}$

Now substituting the values into the formula,

 $\begin{array}{c}\text{n}=\frac{\text{P}\text{V}}{\text{R}\text{T}}\end{array}$

  $\begin{array}{c}=\frac{0.990\times 12}{0.0821\times 623}\end{array}$

  $\begin{array}{c}=0.232\text{m}\text{o}\text{l}\end{array}$

The molar mass of water is 18 g/mol.

Mass of water vapor is given by:

 $\begin{array}{c}\text{m}=\text{M}\times \text{n}\end{array}$

  $\begin{array}{c}\text{=18×0}\text{.232}\end{array}$

  $\begin{array}{c}\text{=4}\text{.176}\text{g}\end{array}$

Conclusion

  $\text{4}\text{.176 g}$ water vapor is produced .