What is the mass of potassium chlorate required to liberate 3.36 dm 3 of oxygen at STP? [Molar mass of potassium chlorate = 122.5 g.mol − 1 ]

Question

TestHub · Accepted Answer

The balanced chemical equation for the decomposition of potassium chlorate is:

$2 K Cl O_{3} \to 2 K Cl + 3 O_{2}$

Number of moles of oxygen gas ( $O_{2}$ ) produced at STP is given by:

$n_{O_{2}} = \frac{V}{22.4} = \frac{3.36}{22.4} = 0.15 mol$

From the balanced equation, $3 mol$ of $O_{2}$ is produced from $2 mol$ of $K Cl O_{3}$ .

Number of moles of $K Cl O_{3}$ required is:

$n_{K Cl O_{3}} = \frac{2}{3} \times n_{O_{2}} = \frac{2}{3} \times 0.15 = 0.10 mol$

Mass of $K Cl O_{3}$ required is:

$m = n_{K Cl O_{3}} \times Molar mass = 0.10 \times 122.5 = 12.25 g$

Answer: $12.25$ g.

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