Chapter 13, Problem 89A

Interpretation Introduction

Interpretation:

The volume of oxygen gas produced by $20.8g$ of potassium chlorate needs to be determined.

Concept introduction:

Stoichiometry of the reaction can be used to calculate the result.

Answer to Problem 89A

$20.8gKCl{O}_{3}willproduce5.70L{O}_{2}atSTP$

Explanation of Solution

The given question involves the chemical equation as:

$2KCl{O}_3\to 2KCl+3O_2$

According to Avogadro’s principle,

$\begin{array}{l}2molKCl{O}_{3}produce3\times 22.4L{O}_{2}atSTP\end{array}$

$\begin{array}{c}MolecularmassofKCl{O}_3=39+35.5+3\times 16\\ =122.5g\end{array}$

$\begin{array}{l}245g(2\times 122.5g)KCl{O}_{3}willproduce67.2L{O}_2\left(3\times 22.4\right)atSTP\\ \therefore 20.8gKCl{O}_{3}willproduce5.70L{O}_2\left(20.8\times \frac{67.2}{245}\right)atSTP\end{array}$

Conclusion

$20.8gKCl{O}_{3}willproduce5.70L{O}_{2}atSTP$