What volume of oxygen, measured at standard conditions, can be produced by heating 10 g of potassium chlorate $K C l O_{3}$ ?

7.24 L

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5.47 L

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4.27 L

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2.74 L

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Solution

The correct option is D $2.74 L$
$2 K C l O_{3} \to 2 K C l + 3 O_{2}$

Moles of $K C l O_{3} = \frac{10}{122.5}$

Moles of $O_{2} = \frac{3}{2} \times \frac{10}{122.5}$

Volume of $O_{2} = \frac{3}{2} \times \frac{10}{122.5} \times 22.4 = 2.74 l i t r e s$

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