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ICSE Class 10 Chemistry 2013 Paper

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O2\mathrm{O}_2 is evolved by heating KClO3\mathrm{KClO}_3 using MnO2\mathrm{MnO}_2 as a catalyst
2KClO3→MnO22KCl+3O22 \mathrm{KClO}_3 \xrightarrow[\mathrm{MnO}_2]{} 2 \mathrm{KCl} + 3 \mathrm{O}_2
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Question : 77 of 84
Marks: +1, -0
Calculate the mass of KClO3\mathrm{KClO}_3 required to produce 6.72 litre of O2\mathrm{O}_2 at STP. [atomic masses of K=39,Cl=35.5\mathrm{K}=39, \mathrm{Cl}=35.5 , O=16]\mathrm{O}=16]
Solution:  
2KClO3→MnO22KCl+3O22×(39+35.5+3×16)3×22.4=2×122.5=245g=67⋅2l\begin{array}{lllll} 2 \mathrm{KClO}_3 & \xrightarrow{\mathrm{MnO}_2} & 2 \mathrm{KCl} & + & 3 \mathrm{O}_2 \\ 2 \times (39+35.5+3 \times 16) & & & & 3 \times 22.4 \\ =2 \times 122.5=245 \mathrm{g} & & & & =67 \cdot 2 \mathrm{l} \end{array}
67.2l67.2 \mathrm{l} of O2\mathrm{O}_2 produced by KClO3\mathrm{KClO}_3 at STP
=245g=245 \mathrm{g}
6.72l6.72 \mathrm{l} of O2\mathrm{O}_2 produced by KClO3\mathrm{KClO}_3
  =  245×6.7267.2\;=\;\frac{245 \times 6.72}{67.2}
  =24.5g\;=24.5 \mathrm{g}
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