# Question 30233

Apr 29, 2015

The student dissolved 0.30 moles of iron (III) sulfate.

Because iron (III) sulfate is soluble in aqueous solution, it will dissociate into $F {e}^{3 +}$ and $S {O}_{4}^{-}$ ions according to the balanced chemical equation

$F {e}_{2} {\left(S {O}_{4}\right)}_{3 \left(a q\right)} \to \textcolor{red}{2} F {e}_{\textrm{\left(a q\right]}}^{3 +} + 3 S {O}_{4 \left(a q\right)}^{2 -}$

Notice that have a $1 : \textcolor{red}{2}$ mole ratio between $F {e}_{2} {\left(S {O}_{4}\right)}_{3}$ and $F {e}^{3 +}$, which means that 1 mole of iron (III) sulfate produces 2 moles of iron (III) ions in solution.

You can use the molarity of the ions to determine how many moles were produced

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{F {e}^{3 +}} = \text{0.60 M" * "1.000 L" = "0.60 moles}$ $F {e}^{3 +}$

Therefore, the number of moles of iron (III) sulfate needed to produce this many moles of iron (III) ions will be

0.60cancel("moles"Fe^(3+)) * ("1 mole"Fe_2(SO_4)_3)/(color(red)(2)cancel("moles"Fe^(3+))) = color(green)("0.30 moles"# $\textcolor{g r e e n}{F {e}_{2} {\left(S {O}_{4}\right)}_{3}}$