Question

How many moles of potassium chlorate are heated to produce \(11\cdot 2\) L oxygen at STP ?

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

Hint:

Write the decomposition equation and use 22.4 L per mole at STP.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Heating potassium chlorate decomposes it into potassium chloride and oxygen. We need the moles of \(\text{KClO}_3\) that give \(11.2\) L of \(\text{O}_2\) at STP.
At STP one mole of any gas occupies \(22.4\) L.

Step 2: Key Formula or Approach:
\[ 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \]
So 2 mol of \(\text{KClO}_3\) give 3 mol of \(\text{O}_2\).

Step 3: Detailed Explanation:
Moles of \(\text{O}_2\) formed:
\[ n(\text{O}_2) = \frac{11.2}{22.4} = 0.5 \text{ mol} \]
From the equation, \(n(\text{KClO}_3) = \frac{2}{3}\times n(\text{O}_2)\).
\[ n(\text{KClO}_3) = \frac{2}{3}\times 0.5 = \frac{1}{3}\text{ mol} \]
The value \(\frac{1}{2}\) would come from a 1:1 ratio and \(\frac{2}{3}\) from inverting the ratio, so neither fits the balanced equation.

Final Answer:
Moles of potassium chlorate needed are one third of a mole, option (B). \[ \boxed{\frac{1}{3}\text{ mol}} \]