O₂ is evolved by heating KClO₃ using MnO₂ as a catalyst.

2KClO₃ $\xrightarrow{\text{MnO}_2}$ 2KCl + 3O₂

(a) Calculate the mass of KClO₃ required to produce 6.72 litre of O₂ at STP.

(atomic masses of K = 39, Cl = 35.5, 0 = 16).

(b) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.

(c) Calculate the volume occupied by 0.01 mole of CO₂ at STP.

$\begin{matrix} 2\text{KClO}$3 & \xrightarrow{MnO2} & 2\text{KCl} & + & 3\text{O}_2 \ 2[39 + 35.5 & & 2[39 & & 3(22.4) \ + 3(16)] & & + 35.5] & & = 67.2 \text{ lit.} \ = 245 \text{ g} & & = 149 \text{g} \ \end{matrix}

(a)

67.2 lit. of O₂ is produced by 245 g of KClO₃

∴ 6.72 lit of O₂ will be obtained from $\dfrac{245}{67.2}$ x 6.72 = 24.5 g

(b)

22.4 lit = 1 mole
∴ 6.72 lit = $\dfrac{1}{22.4}$ x 6.72 = 0.3 moles

1 mole = 6.023 x 10²³ molecules
∴ 0.3 moles = 0.3 x 6.023 x 10²³ = 1.806 x 10²³ molecules.

(c)

Volume occupied by 1 mole of CO₂ at STP = 22.4 litres

So, volume occupied by 0.01 mole of CO₂ at STP
= 22.4 × 0.01
= 0.224 litres