Chapter 13.3, Problem 43PP

Interpretation Introduction

Interpretation:

The amount of carbon dioxide produced by $2.38Kg$ of calcium carbonate is to be determined.

Concept introduction:

Stochiometry concerns are used to calculate the result.

Answer to Problem 43PP

$533.12LC{O}_{2}isproducedby2.38KgCaC{O}_3$

Explanation of Solution

The given question involves the chemical equation as:

$CaC{O}_3\to CaO+C{O}_2$

According to Avogadro’s principle,

$1molCaC{O}_{3}willproduce22.4LC{O}_{2}atSTP$

Now to calculate volume of carbon dioxide in given amount of calcium carbonate, amount is to be converted into moles.

Molar mass of calcium carbonate is 100.

$\begin{array}{c}n=\frac{m}{M}\\ =\frac{2.38\times {10}^3}{100}\\ =23.8mol\end{array}$

Now, $1molCaC{O}_{3}produces22.4LC{O}_{2}atSTP$

$23.8molCaC{O}_{3}willproduce533.12LC{O}_2(23.8\times 22.4)atSTP$

Conclusion

$2.38KgCaC{O}_{3}produces533.12LC{O}_2$