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

`\large \boxed{\text{2.012 mol AlCl}_(3)}`

Step-by-step 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) * \frac{\text{2 mol AlCl}_(3)}{\textbf{3 mol Cl}_(2)}\\\\= \text{2.012 mol AlCl}_(3)\\\text{The reaction produces $\large \boxed{\textbf{2.012 mol AlCl}_(3)}$}`