Question

200 g of impure sample of KClO3 on heating gives
48 g of 02
2KCIO3 - 2KCI + 302
The percentage purity of KClO3 is (Molar mass of
KCIO3 = 122.5 g mol-1)​

Answer 1

Answer:

The answer is in the below image

Explanation:

Answer 2

Concept Introduction: Percentage Purity is one of the basic concepts of Chemistry.

Given:

We have been Given: 200 g of impure sample of KClO3 on heating gives

48 g of 02

2KCIO3 - 2KCI + 302

(Molar mass of KCIO3 = 122.5 g mol-1)

To Find:

We have to Find: The percentage purity of KClO3.

According to the problem,

2KCIO3 - 2KCI + 302

Molar mass of 2 molecules of KCLO3 is

$245g$

Molar Mass of 3 molecules of O2 is

$96g$

So to produce

$48g$

of O2, KCLO3 needed is

$122.5g$

Now, in

$200g$

of Impure KCLO3,

$122.5g$

is pure.

therefore, in

$100g$

of impure KCLO3, the percentage purity is

$\frac{122.5 \times 100}{200} = \frac{1225}{20} = 61.25\%$

Final Answer: The Percentage Purity of KCLO3 is

$61.25\%$

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