Question

Find the mass of potassium chlorate required to liberate $5.6 \text{ dm}^3$ of oxygen gas at STP? (molar mass of $\text{KClO}_3 = 122.5\text{ g/mol}$ )}

• A. $12.25\text{ g}$
• B. $15.32\text{ g}$
• C. $20.40\text{ g}$
• D. $49.00\text{ g}$

Hint:

Always balance the chemical equation before performing stoichiometric calculations.

Answer 1

The Correct Option is C

Step 1: Concept The balanced decomposition reaction is: $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$.

Step 2: Meaning 2 moles of $KClO_3$ ($2 \times 122.5\text{ g}$) produce 3 moles of $O_2$ ($3 \times 22.4 \text{ dm}^3$ at STP).

Step 3: Analysis $67.2 \text{ dm}^3$ of $O_2$ is produced by $245\text{ g}$ of $KClO_3$.
So, $5.6 \text{ dm}^3$ of $O_2$ is produced by $\frac{245 \times 5.6}{67.2}\text{ g}$.

Step 4: Conclusion Mass required $= 245 / 12 = 20.416\text{ g} \approx 20.40\text{ g}$. Final Answer: (C)