A sample of ${K C l O}_{3}$ on decomposition yielded 448 mL of oxygen gas at NTP.
Calculate (i) Weight of oxygen product (ii) Weight of ${K C l O}_{3}$ originally taken (iii) Weight of $K C l$ produced.

${2 K C l O}_{3} Δ - \to {2 K C l + 3 O}_{2}$

$[K = 39, C = 25.5$ and $O = 16]$

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Solution

$2 K C l O_{3} Δ - \to 3 O_{2} + 2 K C l 2 m o l e 3 m o l e 2 m o l e$

Number of moles of $O_{2}$ at S.T.P.= $\frac{V (m l)}{22400} = \frac{448}{22400} = 0.02 m o l e$

Mass of 1 mole of $O_{2}$ = 32 grams

Mass of 0.02 mole of $O_{2}$ =32 $\times$ 0.02 = 0.64 grams

Mass of $K C l O_{3}$ =moles(n) $\times$ molecular weight=2 $\times$ 122.5 = 245 grams

Weight of $K C l O_{3}$ to produce 3 moles of $O_{2}$ = 245 grams

Weight of $K C l O_{3}$ to produce 0.02 moles of $O_{2}$ = $\frac{245 \times 0.02}{3} =$ 1.63 grams

Mass of $K C l$ produced = Mass of $K C l O_{3}$ -Mass of $O_{2}$ produced $= 1.63 - 0.64 = 0.993$ grams

Weight of $O_{2}$ produced = 0.64 grams

Weight of $K C l O_{3}$ originally taken = 1.63 grams

Weight of $K C l$ produced = 0.993 grams

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