Question

how many moles of potassium chlorate to be heated to produce 11.2 litre oxygen ?

Answer 1

Answer on Question #45402 – Chemistry – Other

Question

How many moles of potassium chlorate to be heated to produce 11.2 litre oxygen?

Solution

Number of moles in 11.2 litres of oxygen:

$$n_{O_2} = \frac{V_{O_2}}{V_m} = \frac{11.2 \, L}{22.4 \, \frac{L}{mol}} = 0.5 \, mol,$$

where $V_m$ – molar volume.

The balanced chemical equation of potassium chlorate thermal decomposition is as follows

$$2 \, \mathrm{KClO_3}(s) \rightarrow 3 \, \mathrm{O_2}(g) + 2 \, \mathrm{KCl}(s)$$

As is clear from the chemical equation, 2 moles of $\mathrm{KClO_3}$ are to be heated to produce 3 moles of $\mathrm{O_2}$, i.e. molar ratio $n_{\mathrm{KClO_3}} / n_{\mathrm{O_2}} = 2 / 3$. Hence

$$n_{\mathrm{KClO_3}} = \frac{2}{3} \cdot n_{\mathrm{O_2}} = \frac{2 \cdot 0.5}{3} = 0.33 \, mol$$

Answer: about 0.33 mole of potassium chlorate is to be heated to produce 11.2 litres of oxygen.

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