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: Understanding the Concept:
Use stoichiometry and molar volume at STP to relate volume of gas to mass of reactant.
Step 2: Key Formula or Approach:
Moles of gas $= \text{Vol at STP} / 22.4\text{ L}$.
Balanced equation: $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$.
Step 3: Detailed Explanation:
- Moles of $\text{O}_2 = 5.6 / 22.4 = 0.25\text{ mol}$.
- From reaction, 3 mol $\text{O}_2$ comes from 2 mol $\text{KClO}_3$.
- So, $0.25$ mol $\text{O}_2$ comes from $(2/3) \times 0.25 = 1/6\text{ mol } \text{KClO}_3$.
- Mass $= \text{moles} \times \text{molar mass} = (1/6) \times 122.5 = 20.416\text{ g}$.
Approximated to $20.40\text{ g}$.
Step 4: Final Answer:
Mass required is $20.40\text{ g}$.