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MHT CET · Chemistry · Redox Reactions

What is the mass of \(\mathrm{KClO}_{3(\mathrm{~s})}\) required to liberate 22. \(4 \mathrm{dm}^3\) oxygen at STP during thermal decomposition? (Molar Mass of \(\mathrm{KClO}_{3(\mathrm{~s})}=122.5 \mathrm{~g} / \mathrm{mol}\) )

  1. A \(122.5 \mathrm{~g}\)
  2. B \(81.67 \mathrm{~g}\)
  3. C \(10.25 \mathrm{~g}\)
  4. D \(8.16 \mathrm{~g}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(81.67 \mathrm{~g}\)

Step-by-step Solution

Detailed explanation

\(
\begin{aligned}
& 2 \mathrm{KClO}_3 \\
& {[2 \mathrm{moles}]}
\end{aligned} \longrightarrow 2 \mathrm{KCl}+\underset{[3 \mathrm{moles}]}{3 \mathrm{O}_2 \uparrow}
\)
[3 moles]
\(
2 \text { moles of } \mathrm{KClO}_3=2 \times 122.5=245 \mathrm{~g}
\)
3 moles of \(\mathrm{O}_2\) at STP occupy \(=\left(3 \times 22.4 \mathrm{dm}^3\right)\).
Thus, \(245 \mathrm{~g}\) of potassium chlorate will liberate \(67.2 \mathrm{dm}^3\) of oxygen gas.
Let ' \(x\) ' gram of \(\mathrm{KClO}_3\) liberate \(22.4 \mathrm{dm}^3\) of oxygen gas at S.T.P.
\(
\therefore \quad x=\frac{245 \times 22.4}{3 \times 22.4}=81.67 \mathrm{~g}
\)