What is the volume of oxygen liberated at STP when 12.25 g of potassium chlorate is subjected to heating?

4.2 L

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1.68 L

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2.24 L

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3.36 L

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Solution

The correct option is D 3.36 L
The chemical equation for decomposition of $K C l O_{3}$ can be written as:

$2 K C l O_{3} \to 2 K C l + 3 O_{2}$

$245$ g of $K C l O_{3}$ gives $67.2$ L of $O_{2}$ at STP.

$12.25$ g of $K C l O_{3}$ gives $\frac{67.2}{245} \times 12.25$ $= 3.36$ L.

Hence, option D is correct.

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What is the volume of oxygen liberated at STP when 12.25 g of potassium chlorate is subjected to heating?

The correct option is D 3.36 LThe chemical equation for decomposition of KClO3 can be written as:2KClO3→2KCl+3O2245 g of KClO3 gives 67.2 L of O2 at STP.12.25 g of KClO3 gives 67.2245×12.25 =3.36 L.Hence, option D is correct.