Chapter 13.3, Problem 45PP

Interpretation Introduction

Interpretation:

The volume of carbon dioxide produced by $28g$ of sodium bicarbonate in the given chemical reaction needs to be determined.

Concept introduction:

Stochiometry concerns are used to calculate the result.

Answer to Problem 45PP

$7.37LC{O}_{2}isproducedwith28gNaHC{O}_3$

Explanation of Solution

The given question involves the chemical equation as:

$NaHC{O}_3+C{H}_{3}COOH\to C{H}_{3}COONa+C{O}_2+H_{2}O$

According to Avogadro's principle,

$\begin{array}{l}84gNaHC{O}_{3}produce22.4LC{O}_{2}atSTP\end{array}$

• $\begin{array}{l}28gNaHC{O}_{3}willproduce7.5LC{O}_2\left(28\times \frac{22.4}{84}\right)atSTP\end{array}$

But temperature conditions are being changed. So applying Charles' Law-

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Or

${\mathrm{V}}_2=\frac{T_2{V}_1}{T_1}$

Given data is:

${\text{V}}_1=7.5L$

${\text{T}}_1=25°C$

${\text{T}}_2=20°C$

$\begin{array}{l}{T}_{}(K)=T_{}(°C)+273\\ T_1(K)=25+273\\ =298K\\ T_2(K)=20+273\\ =293K\end{array}$

On substituting the values in the formula:

$\begin{array}{c}{\mathrm{V}}_2=\frac{293\times 7.5}{\begin{array}{l}298\end{array}}\\ =7.37L\end{array}$

Therefore new volume is $7.37L$

Conclusion

$28gNaHC{O}_{3}produces7.37LC{O}_{2}at20°C$