MHT CET · Chemistry · Some Basic Concepts of Chemistry
In the reaction
\(2 \mathrm{KClO}_{3(\mathrm{~s})} \longrightarrow 2 \mathrm{KCl}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})} \Delta \mathrm{H}^{\circ}=-78 \mathrm{~kJ} .\) If \(33 \cdot 6 \mathrm{~L}\) of oxygen gas is liberated at S.T.P. What is the mass of \(\mathrm{KCl}_{(s)}\) produced \(?\) (at. mass \(\mathrm{K}=39, \mathrm{Cl}=35 \cdot 5 \mathrm{~g} \mathrm{~mol}^{-1}\) )
- A \(48 \cdot 0 \mathrm{~g}\)
- B \(7.45 \mathrm{~g}\)
- C \(24 \cdot 0 \mathrm{~g}\)
- D \(74.5 \mathrm{~g}\)
Answer & Solution
Correct Answer
(A) \(48 \cdot 0 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
\(2 \mathrm{KClO}_{3(s)} \longrightarrow 2 \mathrm{KCl}_{(s)}+3 \mathrm{O}_{2(s)} \)
\( 2 \times 74.5 \mathrm{~g} 3 \times 22.4 \mathrm{~L} \)
\(=149 \mathrm{~g} \quad=67.2 \mathrm{~L}\)
Now, \(67.2 \mathrm{~L}\) of \(\mathrm{O}_{2}=149 \mathrm{~g}\) of \(\mathrm{KCl}\) at \(\mathrm{STP}\)
\(33.6 \mathrm{~L}\) of \(\mathrm{O}_{2}=\mathrm{x} \mathrm{g}\) of \(\mathrm{KCl}\)
\(\therefore x=\frac{149 \times 33.6}{67.2}=74.5 \mathrm{~g}\) of \(\mathrm{KCl}\)
\( 2 \times 74.5 \mathrm{~g} 3 \times 22.4 \mathrm{~L} \)
\(=149 \mathrm{~g} \quad=67.2 \mathrm{~L}\)
Now, \(67.2 \mathrm{~L}\) of \(\mathrm{O}_{2}=149 \mathrm{~g}\) of \(\mathrm{KCl}\) at \(\mathrm{STP}\)
\(33.6 \mathrm{~L}\) of \(\mathrm{O}_{2}=\mathrm{x} \mathrm{g}\) of \(\mathrm{KCl}\)
\(\therefore x=\frac{149 \times 33.6}{67.2}=74.5 \mathrm{~g}\) of \(\mathrm{KCl}\)
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