# Question f78be

Approx. $90 \cdot g$ chlorine gas are required.
$F e + \frac{3}{2} C {l}_{2} \rightarrow F e C {l}_{3}$
$\text{Moles of ferric chloride}$ $=$ $\frac{140 \cdot g}{162.2 \cdot g \cdot m o {l}^{-} 1} = 0.863 \cdot m o l .$
And given the stoichiometry we need $\frac{3}{2} \times 2 \times 35.45 \cdot g \cdot m o {l}^{-} 1 \times 0.863 \cdot m o l$ $=$ ?*g# $\cdot C {l}_{2}$