Question

O₂ is evolved by heating KCIO₃ using MnO₂ as a catalyst.
\[\ce{2KCI3 ->[MnO2] 2KCl + 3O2}\]

1. Calculate the mass of KClO₃ required to produce 6.72 litres of O₂ at STP. [atomic masses of K = 39, Cl = 35.5, O = 16].
2. Calculate the number of moles of oxygen present in the above volume and also the number of molecules.
3. Calculate the volume occupied by 0.01 mole of CO₂ at STP.

Answer 1

2KClO3	\[\ce{->[MnO2]}\]	2KCl	+	3O2
2[39 + 35.5 + 48]g		2[39 + 35.5]g		3[22.4]lt
245 g		67.2 lt

(i) 3 Volumes of Oxygen require KClO₃ = 2 Volumes

So, 1 Vol. of oxygen will require KClO₃ = `2/3` volumes

So, 6.72 litres of Oxygen will require KClO₃ = `2/3 xx 6.72`

= 4.48 litres

22.4 litres of KClO₃ has mass = 122.5g

∴ 4.48 litres of KClO₃ will have mass = `[122.5]/[22.4] xx 4.48`

= 24.5 g

(ii) 22.4 litres of oxygen = 1 mole

So, 6.72 litres of oxygen = `6.72/22.4`

= 0.3 moles

No. of molecules present in 1 mole of O₂ = 6 × 10²³

So, no. of molecules present in 0.3 mole of O₂

= `(6 xx 10^23 xx 0.3)/1`

= 1.8 × 10²³ 

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

So,volume occupied by 0.01 mole of CO₂ at STP = `(22.40 xx 0.01)/1`

= 0.224 litres

= 224 cm³