Question

How much potassium chlorate should be heated to produce 2.24L of oxygen at NTP?

Answer 1

The equation is : ${2KClO}_3\to 2KCl+{3O}_2$

$2moles\to 3moles$

i.e. $2(39+35.5+48)\to 3\times 22.4$ of oxygen

$245$g of ${KClO}_3$ $3\times 22.4$lit

i.e. $3\times 22.4$ lit of $O_2\to 245g$ of ${KClO}_3$

$\therefore 2.24$ lit of $O_2\to \frac{245}{3\times 22.4}\times 2.24$

$=8.17g$