# How many moles of chlorine gas will be required to react with sufficient iron to produce 14 moles of iron (III) chloride?

## $2 F e \left(s\right) + 3 C {l}_{2} \left(g\right) \to 2 F e C {l}_{3} \left(g\right)$

The reaction is $2 F e + 3 C {l}_{2} \to 2 F e C {l}_{3}$
So the mol ratio $C {l}_{2} \div F e C {l}_{3} = 3 \div 2$
So for 14 mols of $F e C {l}_{3}$ we need $\frac{3}{2} \times 14 = 21$ mols of $C {l}_{2}$