Chapter 13, Problem 107A

Interpretation Introduction

Interpretation:

Volume of CO₂per gram is to be calculated for the given reactions at given conditions of temperature and pressure.

Concept introduction:

Ideal gas equation and volume-volume relationship is used to calculate the result

Answer to Problem 107A

Volume of CO₂per gram in reaction first is 0.24 L/gram and second is 0.48 L/gram

Explanation of Solution

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

$\mathrm{PV}=nRT$

Or

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

Given data for CO₂is:

$\text{n=1}mol$

$\mathrm{P}=0.985atm$

$\text{R=}0.0821Latm{K}^{-1}mo{l}^{-1}$

$\begin{array}{c}\mathrm{T}=210°C\\ =210+273\\ =483K\end{array}$

Now substituting the values into formula,

$\begin{array}{l}V=\frac{nRT}{P}\\ =\frac{1\times 0.0821\times 483}{0.985}\\ =40.2LC{O}_2\\ \begin{array}{l}Reaction1:\hfill \\ {\text{Molar ratio of NaHCO}}_{\text{3}}{\text{and CO}}_{\text{2}}\text{is 2:1}\hfill \end{array}\\ 2molofNaHC{O}_{3}produces40.2LC{O}_2 \\ 168g/molNaHC{O}_{3}produces40.2LC{O}_2\\ 1gramwillproduce=\frac{40.2}{168}\\ =0.24L/gram\\ \begin{array}{l}Reaction2:\hfill \\ {\text{Molar ratio of NaHCO}}_{\text{3}}{\text{and CO}}_{\text{2}}\text{is 1:1}\hfill \end{array}\\ 1molofNaHC{O}_{3}produces40.2LC{O}_2 \\ 84g/molNaHC{O}_{3}produces40.2LC{O}_2\\ 1gramwillproduce=\frac{40.2}{84}\\ =0.48L/gram\end{array}$

Conclusion

Volume of CO₂per gram in reaction first is 0.24 L/gram and second is 0.48 L/gram