Question

1.5 g of sample of KClO3 on heating decomposed in two ways
2KlO3---->2KCl + 3O2
4 KClO3 ----->3KClO4 + KCl
if the amount of KClO4 present in the residue is 0.395g.the volume of O2 collected at STP is .....?......ml

Answer 1

Answer:

276.8 litres

Explanation:

Molecular weight of KClO₄ = 138.55

Molecular weight of KClO₃ = 122.55

moles of KClO₄ = weight in grams/molecular weight = 0.395/138.55 = 0.003

4 moles of KClO₃ gives 3 moles of KClO₄

Therefore, 0.004 moles of  KClO₃ will give 0.003 moles of KClO₄

Moles of KClO₃ initially present = 1.5/122.55 = 0.0122

Moles of KClO₃ used in the first reaction = 0.0122 - 0.004 = 0.00823

2 moles of KClO₃ produces 3 moles of O₂

Therefore, 0.00823 moles of KClO₃ will produce

= 3/2 × 0.00823 moles of O₂

= 0.0124 moles of O₂

At STP, volume of 1 mole of gas = 22.4 litre

Therefore volume of 0.0124 moles of O₂ = 22.4 × 0.0124 litres

                                                                    = 0.2768 litres

                                                                    = 276.8 litres

Answer 2

here's the answer and hope this helps you