Chapter 13, Problem 85A

Interpretation Introduction

Interpretation: The mass of water vapor produced by $6.5\text{L}$ hydrogen sulfide gas needs to be determined.

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

Answer to Problem 85A

The mass of water vapor produced is $\text{9}\text{.72}\text{g}$ .

Explanation of Solution

The given equation is:

  $2{\text{H}}_2\text{S}+{\text{O}}_2\to 2{\text{H}}_2\text{O}+2\text{S}$

Equation shows $2\text{L}$ of hydrogen sulfide forms $2\text{L}{\text{H}}_2\text{O}$

Therefore $6.5\text{L}{\text{H}}_2\text{S}$ will be form water vapor:

  $\begin{array}{c}6.5\text{L}{\text{H}}_2\text{S}=6.5\text{L}{\text{H}}_2\text{S}\times \frac{2\text{L}{\text{H}}_2\text{O}}{2\text{L}{\text{H}}_2\text{S}}\end{array}$

      $\begin{array}{c}=6.5\text{L}{\text{H}}_2\text{O}\end{array}$

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:

  $\text{V}=6.5\text{L}$

  $\text{P}=2\text{a}\text{t}\text{m}$

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

  $\text{T=290}\text{K}$

Now substituting the values into the formula,

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

  $\begin{array}{c}=\frac{2\times 6.5}{0.0821\times 290}\end{array}$

  $\begin{array}{c}=0.54\text{m}\text{o}\text{l}\end{array}$

The molar mass of water is 18 g/mol.

Mass of water vapor is given by:

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

  $\begin{array}{c}\text{=18×0}\text{.54}\end{array}$

  $\begin{array}{c}\text{=9}\text{.72}\text{g}\end{array}$

Conclusion

  $\text{9}\text{.72}\text{g}$ water vapor is produced .