# Question #62d92

Mar 28, 2015

A very important aspect to keep in mind here - you don't actually have 4 $\text{So}$ elements on the reactants' side and 12 $\text{So}$ elements on the products' side because $\text{So}$ $\textcolor{red}{\text{is not an element}}$.

You're dealing with the sulfate anion, $S {O}_{4}^{2 -}$, which is comprised of 1 sulfur atom, $\text{S}$, and 4 oxygen atoms, $\text{O}$.

This means that you can try to balance the equation either by looking at individual elements, like the other answer attempts to do, or by taking the sulfate anion as a group.

The latter options implies that you have 2 hydrogen atoms, 1 sulfate group, and 1 iron atom on the left side of the equation, and 2 iron atoms, 2 hydrogen atoms, and 3 sulfate groups on the right side of the equation.

${H}_{2} S {O}_{4} + F e \to F {e}_{2} {\left(S {O}_{4}\right)}_{3} + {H}_{2}$

Balance the sulfate group first and focus on the rest of the species afterwards. So, you have three times more sulfate groups on the right hand side of the equation $\to$ multiply the compound that contains the sulfate group by 3 on the left hand side.

$\textcolor{red}{3} {H}_{2} S {O}_{4} + F e \to F {e}_{2} {\left(S {O}_{4}\right)}_{\textcolor{red}{3}} + {H}_{2}$

By multiplying this compound by 3, you've increased the number of hydrogen atoms on the left hand side to 6 $\to$ multiply the hydrogen on the right hand side by 3 to balance them out.

$\textcolor{red}{3} {H}_{\textcolor{b l u e}{2}} S {O}_{4} + F e \to F {e}_{2} {\left(S {O}_{4}\right)}_{\textcolor{red}{3}} + \textcolor{red}{3} {H}_{\textcolor{b l u e}{2}}$

Finally, you've got 2 iron atoms on the right hand side $\to$ multiply the iron by 2 on the left hand side to balance them out

$\textcolor{red}{3} {H}_{\textcolor{b l u e}{2}} S {O}_{4} + \textcolor{g r e e n}{2} F e \to F {e}_{\textcolor{g r e e n}{2}} {\left(S {O}_{4}\right)}_{\textcolor{red}{3}} + \textcolor{red}{3} {H}_{\textcolor{b l u e}{2}}$