One mole of potassium chlorate $\left({\mathrm{KClO}}_3\right)$ is thermally decomposed and excess of aluminium is burnt in the gaseous product. How many mol of aluminium oxide $\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)$ are formed?

When one mole of potassium chlorate (KClO₃) undergoes thermal decomposition, it produces oxygen gas (O₂), which then reacts with excess aluminum (Al) to form aluminum oxide (Al₂O₃). The reactions involved are:

Decomposition of KClO₃:

2 KClO₃ → 2 KCl + 3 O₂

From this equation, 2 moles of KClO₃ produce 3 moles of O₂.

Therefore, 1 mole of KClO₃ will produce:

3/2 moles of O₂ = 1.5 moles of O₂

Formation of Al₂O₃:

4 Al + 3 O₂ → 2 Al₂O₃

Here, 3 moles of O₂ react with aluminum to produce 2 moles of Al₂O₃.

Since we have 1.5 moles of O₂ (which is half of 3 moles), it will produce half the amount of Al₂O₃:

2/2 moles of Al₂O₃ = 1 mole of Al₂O₃

Correct Answer: a) 1

This means that 1 mole of KClO₃, when decomposed, produces sufficient oxygen to form 1 mole of aluminum oxide when reacted with excess aluminum.