Question

How much potassium chlorate is needed to obtain 2.4 litre oxygen at NTP ?

Answer 1

$2KCl{O}_3⟶2KCl+3O_2$

$2\times 22.4LatNTP$ $3\times 22.4LatNTP$

$\because 2\times 22.4L$ of $KCl{O}_3$ generates $3\times 22.4L$ of $O_2$ at NTP.
$\therefore 2.4L$ of $O_2$ needs, $=\frac{2\times 22.4\times 2.4}{3\times 22.4}=1.6L$ of $KCl{O}_3$