Question #62d92

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Question #62d92

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Answer 1 · 2015-03-28T12:05:53

A very important aspect to keep in mind here - you don't actually have 4 $So$ $"So"$ elements on the reactants' side and 12 $So$ $"So"$ elements on the products' side because $So$ $"So"$ $is not an element$ $color(red)("is not an element")$ .

You're dealing with the sulfate anion, $S O_{4}^{2 -}$ $SO_4^(2-)$ , which is comprised of 1 sulfur atom, $S$ $"S"$ , and 4 oxygen atoms, $O$ $"O"$ .

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.

$H_{2} S O_{4} + F e \to F e_{2} {(S O_{4})}_{3} + H_{2}$ $H_2SO_4 + Fe -> Fe_2(SO_4)_3 + H_2$

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 $\to$ $->$ multiply the compound that contains the sulfate group by 3 on the left hand side.

$3 H_{2} S O_{4} + F e \to F e_{2} {(S O_{4})}_{3} + H_{2}$ $color(red)(3)H_2SO_4 + Fe -> Fe_2(SO_4)_(color(red)(3)) + H_2$

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

$3 H_{2} S O_{4} + F e \to F e_{2} {(S O_{4})}_{3} + 3 H_{2}$ $color(red)(3)H_(color(blue)(2))SO_4 + Fe -> Fe_2(SO_4)_(color(red)(3)) + color(red)(3)H_color(blue)(2)$

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

$3 H_{2} S O_{4} + 2 F e \to F e_{2} {(S O_{4})}_{3} + 3 H_{2}$ $color(red)(3)H_(color(blue)(2))SO_4 + color(green)(2)Fe -> Fe_(color(green)(2))(SO_4)_(color(red)(3)) + color(red)(3)H_color(blue)(2)$

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Question #62d92

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