Question

What volume of $0.4MFeC{l}_3$ will contain 600 mg of $F{e}^{3+}$?

Answer 1

In $FeC{l}_3,$ $Fe$ is in $+3$ state.

$600mg$ of $F{e}^{+3}$ contains, $\frac{600g}{56\times 1000}=0.0107moles$ of $F{e}^{+3}$

$\underset{o.4M}{FeC{l}_3}⟶\underset{0.4M}{F{e}^{+3}}+\underset{3\times 0.4M}{3C{l}^{-}}$

as $n=molarity\times volume$

$volume=\frac{n}{M}=\frac{0.0107moles}{0.4moles{1}^{-1}}$

$v=0.02678L=26.78mL$

Thus, $26.78mL$ will contain $600mg$ of $F{e}^{+3}$.