Question

How many moles of potassium chlorate are heated to produce \(22.4 \text{lit}\) of oxygen at STP ?

• A. \(\frac{1}{2} \text{mole}\)
• B. \(\frac{1}{3} \text{mole}\)
• C. \(\frac{1}{4} \text{mole}\)
• D. \(\frac{2}{3} \text{mole}\)

Hint:

Write the decomposition equation and note that 22.4 L of gas at STP is exactly 1 mole.

Answer 1

The Correct Option is D

Step 1: Mass route:
Oxygen at STP: $22.4$ L is $1$ mol, which is $32$ g of $\text{O}_2$.
The reaction $2\text{KClO}_3 \to 2\text{KCl} + 3\text{O}_2$ uses $2 \times 122.5 = 245$ g of $\text{KClO}_3$ to give $3 \times 32 = 96$ g of $\text{O}_2$.

Step 2: Proportion:
For $32$ g of $\text{O}_2$ the mass of $\text{KClO}_3$ needed is $\frac{245}{96} \times 32 = 81.67$ g.
Moles $= \frac{81.67}{122.5} = 0.667 = \frac{2}{3}$.
The other fractions ($\frac12$, $\frac13$, $\frac14$) would need different oxygen yields, so they are not possible.

Final Answer:
The amount of $\text{KClO}_3$ is $\frac{2}{3}$ mole, option (D). \[ \boxed{\frac{2}{3}\ \text{mole}} \]