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Standard XII

Chemistry

Stoichiometric Calculations

1.0 g sample ...

Question

$1.0$ g sample of $K C l O_{3}$ was heated under such conditions that a part of it decomposed according to the equation:
$2 K C l O_{3} \to 2 K C l + 3 O_{2}$ ,
and the remaining underwent change according to the equation:
$4 K C l O_{3} \to + 3 K C l O_{4} + K C l$ .
If the amount of $O_{2}$ evolved was $146.8$ mL at STP, calculate the percentage by weight of $K C l O_{4}$ in the residue.

49.87

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50.87

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49.00

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Solution

The correct option is A $49.87$
$2 K C l O_{3} 245 g \to 2 K C l 149 g + 3 O_{2} 96 g$
$4 K C l O_{3} 490 g \to 3 K C l O_{4} 415.5 g + K C l 74.5$
Given, total weight of the sample $= 1$ g
Let $x$ g of $K C l O_{3}$ be converted to $K C l$ and $O_{2}$ and $(1 - x)$ g be converted to $K C l O_{4}$ and $K C l$ .
Given, $146.8$ mL of $O_{2}$ was evolved at STP.
Weight of $O_{2} = \frac{146.8 \times 32}{22400} = 0.2097$ g
Weight of $K C l O_{3}$ converted to $K C l$ and $O_{2} =$ $x = \frac{245 \times 0.2097}{96} = 0.5352$ g
Weight of $K C l O_{4}$ formed from remaining $K C l O_{3}$ $= \frac{415.5 \times (1 - x)}{490} = 0.3941$ g
Weight of residue $= 1 -$ weight of $O_{2} = 1 - 0.2097$ g
$∴ %$ of $K C l O_{4}$ in the residue $= \frac{0.3941 \times 100}{(1 - 0.2097)} = 49.87 %$

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Similar questions

Q. A 1 g sample of

K C l O_{3}

was heated under such conditions that a part of it decomposed according to the equation.

(i) 2 K C l O_{3} \to 2 K C l + 3 O_{2}

And the remaining underwent change according to the equation.

(i i) 4 K C l O_{3} \to 3 K C l O_{4} + K C l

If the amount of oxygen evolved was 146.8 mL at NTP, calculate the percentage by weight of

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