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
here we see two mole of KClO3 form 3 mole of Oxygen gas .
hence,
1 mole of KClO3 form 1.5 mole O2 gas

(39+35.5+48) g KClO3 form 1.5 mole O2 gas

10 g KClO3 form 1.5 × 22.4/12.25 L O2 gas e.g 2.74 L of O2 gas

so, ℅ purity = 2.24/2.74 × 100 = 81.75 ℅

Answer 2

Answer:

81.65%(approx)

Explanation:

2KClO3 →2KCl + 3O2

here we see two mole of KClO3 form 3 mole of Oxygen gas .

hence,

1 mole of KClO3 form 1.5 mole O2 gas

(39+35.5+48) g KClO3 form 1.5 mole O2 gas

10 g KClO3 form 1.5 × 22.4/12.25 L O2 gas e.g 2.74 L of O2 gas

so, ℅ purity = 2.24/2.74 × 100 = 81.75 ℅

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