# If 113.0 g of iron reacts with 138.0 g of chlorine gas, what is the limiting reactant?

Jul 11, 2017

Both $\text{ferrous chloride}$, $F e C {l}_{2}$, and $\text{ferric chloride}$, $F e C {l}_{3}$, can be made by direct combination of the elements.......

#### Explanation:

Let's assume that it is $F e C {l}_{2}$, i.e.

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

$\text{Moles of iron} = \frac{113.0 \cdot g}{55.8 \cdot g \cdot m o {l}^{-} 1} = 2.04 \cdot m o l$......

$\text{Moles of}$ $C {l}_{2}$ $\text{gas}$ $= \frac{138.0 \cdot g}{70.9 \cdot g \cdot m o {l}^{-} 1} = 1.95 \cdot m o l$......

And since there is a slightly less molar quantity of $C {l}_{2}$ with respect to the metal, then chlorine gas is the reagent in deficiency, i.e. the limiting reactant.