Calculate the mass of potassium chlorate required to liberate 6.72 ${d m}^{3}$ of oxygen at STP, molar mass of potassium chlorate is 112.5 g ${m o l}^{- 1}$ .

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Solution

The balanced chemical equation for the given reaction is:-
$2 K C L O_{3} \to 2 K C L + 3 O_{2}$

$2 m o l K C L O_{3} w i l l p r o d u c e 3 m o l o f O_{2}$

$A t S T P 22.4 L = 1 m o l e o f O_{2}$

$T h e r e f o r e 6.72 d m^{3} (L) = \frac{6.72 L}{22.4 L m o l^{1}} = 0.3 m o l O_{2}$

$s i n c e 3 m o l o f O_{2} i s p r o d u c e d f r o m 2 m o l o f K C L O_{3}$

$T h e r e f o r e 0.3 m o l O_{2} w i l l r e q u i r e \frac{2 \times 0.3}{3}$
$= 0.2 m o l K C L O_{3}$

$M a s s K C L O_{3} c o r r e s p o n d i n g t o o .2 m o l K C L O_{3} = n u m b e r o f m o l e \times m o l a r m a s s$
= $0.2 \times 112.5 g m o l$

= $22.5 g$

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Calculate the mass of potassium chlorate required to liberate 6.72 dm³of oxygen at STP

= 122.5

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