Answer:
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.
