In the reaction, 2KClO₃ $\to$ 2KCl + 3O₂ when 36.75 g of KClO₃ is heated, the volume of oxygen evolved at N.T.P. will be (nearest integer) :

$2{\mathrm{KClO}}_3\to \mathrm{KCl}+3{\mathrm{O}}_2$

At NTP, one mole of gas occupies 22.4 dm³.

245 g of KClO₃ gives 67.2 dm³ $\left(22.4 {\mathrm{dm}}^3\times 3\right)$ of oxygen. Volume of oxygen liberates from 36.75 g of KClO₃ is:

$\mathrm{Volume} \mathrm{of} {\mathrm{O}}_2=\frac{67.2}{245}\times 36.75=10.00 {\mathrm{dm}}^3$