Answer:
[tex]\large \boxed{\text{A. 3.0 mol}}[/tex]
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
2FeCl₃ + 3Mg ⟶ 3MgCl₂ + 2Fe
n/mol: 4.5
2 mol of FeCl₃ react with 3 mol of Mg
[tex]\text{Moles of FeCl}_{3} = \text{4.5 mol Mg}\times \dfrac{\text{2 mol FeCl}_{3}}{\text{3 mol Mg}}= \textbf{3.0 mol Mg}\\\\\text{The reaction will need $\large \boxed{\textbf{3.0 mol of Mg}}$}[/tex]
