Question

(20 pts) When 3.018 moles of chlorine reacts with excess aluminum, how many moles of
aluminum chloride are formed?
2 Al + 3 Cl2 → 2 AICI3:

Answer 1

Answer:

$\large \boxed{\text{2.012 mol AlCl}_{3}}$

Explanation:

             2Al + 3Cl₂ → 2AlCl₃

n/mol:            3.018

The molar ratio is 2 mol AlCl₃:3 mol Cl₂.

$\text{Moles of AlCl}_{3} = \text{ 3.018 mol Cl}_{2} \times \dfrac{\text{2 mol AlCl}_{3}}{\textbf{3 mol Cl}_{2}}\\\\= \text{2.012 mol AlCl}_{3}\\\text{The reaction produces \large \boxed{\textbf{2.012 mol AlCl}_{3}} }$