Question

what volume of oxygen at stp will be obtained by the action of heat on 20g of kclo3 k=39 cl=35.5 o=16

Answer 1

KClO₃(122.5 g )  →  KCl(74.5g) + O₂(32g) 

122.5 g of KClO₃ decomposes into 32 g of O₂

1 g of KClO₃ decomposes into 32/122.5 = 0.26 g of O₂

20 g of KClO₃ decomposes into 5.22 g of O₂

32 g of O₂ is 22.4 litres at STP 

1g of O₂ is 22.4/32 = 0.7 litres at STP 

5.22 g of O₂ is 3.654 lires at STP 

Answer 2

2KClO3 --> 2KCl + 3O2

GMM of KClO3 = 2 x 122.5g

2 × 122.5g KClO3 forms 3 × 22.4L O2

1g KClO3 forms 3 × 22.4/ 2 × 122.5 O2

20g KClO3 forms {(3 × 22.4/ 2 × 122.5) × 20}L O2 = 5.486 L O2