Question

a 10g sample of kclo3 gave on complete combustion, 2.24 L of O2 at NTP . what % purity sample of KClO3(potassium chlorates) ?

Answer 1

2KClO3 ---> 2KCl + 3O2
Therefore, 2 mol of KClO3    gives     3 mol of O2
122.5 g of KClO3     gives     22.4 X 3 L of O2
10 g of KClO3 = 0.081 mol (approx.)
2.24 L of O2 = 0.1 mol
if 3 mol of O2     is obtained from     2 mol of KClO3
0.1 mol of O2     is obtained from     2/3 X 0.1 = 0.06 mol of KClO3
Therefore, amount of pure potassium chlorate in sample = 0.06 mol
% age purity = 0.06 / 0.081 X 100 = 74.07% approx.

Answer 2

Hey mate here is your answer,

1 mol of O2 = 22.4L volume
therefore,
2.24L volume = 0.1 mol of O2

Mole of KClO3 = 10 ÷ 122.5 = 0.081 mol
Mole of O2
3 Mole of O2 ---- 2 mole of KClO3
0.1 mole of O2 --- 0.1×2÷3 mole of KClO3
= 0.06 mol
% purity = 0.06 ÷ 0.081 × 100
= 74.07%

I hope it will help you
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