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5.48 L

Explanation:

We'll begin by calculating the number of mole in 20 g of KClO₃. This can be obtained as follow:

Mass of KClO₃ = 20 g

Molar mass of KClO₃ = 39 + 35.5 + (16×3)

= 39 + 35.5 + 48

= 122.5 g/mol

Mole of KClO₃ =?

Mole = mass / molar mass

Mole of KClO₃ = 20 / 122.5

Mole of KClO₃ = 0.163 mole

Next, the balanced equation for the reaction. This is given below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 0.163 moles of KClO₃ will decompose to produce = (0.163 × 3)/2 = 0.2445 moles of O₂.

Finally, we shall determine the volume of O₂ produced from the reaction. This is can be obtained as illustrated below:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.2445 moles of O₂ will occupy = 0.2445 × 22.4 = 5.48 L at STP.

Thus, 5.48 L O₂ were obtained from the reaction.