Question

11. Calculate the volume of oxygen at NTP that can be produced by heating 12.25 g of potassium chlorate.
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Answer 1

$\fbox{Answer = 3.36\:L}$

$Explanation :-$

$The\:equation \:for\:the\:decomposition \:of\:$ $potassium \:chlorate\:is\: :$

$2KClO3 ---------) 2KCl + 3O2$

$2 \times{122.5} \:=\:245\:g \:\:gives\:3\:mol(3\times{22.4}\:at\:STP)$

$Vol. \:of\:O2\:at\:NTP\:produced\:$

$by\:245g\:KClO3\:=\:3\times{22.4L}$

$Therefore, \:of\:O2\:at\:NTP\:produced\:by$

$12.25g \:KClO3\:=\:$

$= \: \frac{3\times{22.4}}{245} \times{12.25}\:=$

$=\:\fbox{ 3.36L}$