When 0.110 mol of aluminum are allowed to react with an excess of chlorine gas, Cl2, how many moles of aluminum chloride are produced?

2Al + 3Cl2 = 2AlCl3

0.11 moles of Aluminium chloride

Explanation:

Given:

2Al + 3Cl₂ → 2AlCl₃

According to the balanced equation:

2 moles pf Aluminium form 2 moles of Aluminium chloride.

So, 1 mole of Al will form 1 mole of Aluminium chloride.

0.11 mole of Al is present.

So, 0.11 mol of Al will form 0.11 mol of Aluminium chloride.