Question

6.0 g of a sample of potassium chlorate (KCIO3) gave 1.9 g of oxygen on strong heating. What is the percentage purity of the sample ?​

Answer 1

Answer:

80.82 %

Explanation:

2 KClO3 →Δ 2 KCl + 3 O2

2 (39 + 35.5+ 48) KClO3 −−−−− 3 × 32 g oxygen

96 g of oxygen is produced from 245 g of KClO3

1.9 g of oxygen is produced from   = 245/96 × 1.9 = 4.849 g

But the given sample weight is 6 g% purity = Theoretical weight/Actual weight× 100 = 4.849/6 × 100 = 80.82 %

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