Question

Question #62d92

Answer 1

A very important aspect to keep in mind here - you don't actually have 4 `"So"` elements 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)`