3.067 moles
2AlCl₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaCl₂
We are given;
We are required to determine the number of moles of Aluminium hydroxide formed.
From the equation, 3 moles of Ca(OH)₂ reacts to produce 2 moles of Al(OH)₃
Therefore;
Moles of Al(OH)₃ = Moles of Ca(OH)₂ × 2/3
= 4.6 moles × 2/3
= 3.067 moles
Therefore, 3.067 moles of Al(OH)₃ would be produced.