A very important aspect to keep in mind here – you don’t actually have 4 ##”So”## on the reactants’ side and 12 ##”So”## elements on the products’ side because ##”So”## ##color(red)(“is not an element”)##.

You’re dealing with the sulfate anion, ##SO_4^(2-)##, which is comprised of 1 sulfur atom, ##”S”##, and 4 oxygen atoms, ##”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_2SO_4 + Fe -> Fe_2(SO_4)_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 ##->## multiply the compound that contains the sulfate group by 3 on the left hand side.

##color(red)(3)H_2SO_4 + Fe -> Fe_2(SO_4)_(color(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 ##->## multiply the hydrogen on the right hand side by 3 to balance them out.

##color(red)(3)H_(color(blue)(2))SO_4 + Fe -> Fe_2(SO_4)_(color(red)(3)) + color(red)(3)H_color(blue)(2)##

Finally, you’ve got 2 iron atoms on the right hand side ##->## multiply the iron by 2 on the left hand side to balance them out

##color(red)(3)H_(color(blue)(2))SO_4 + color(green)(2)Fe -> Fe_(color(green)(2))(SO_4)_(color(red)(3)) + color(red)(3)H_color(blue)(2)##