Question

How do you balance the equation: `(A)FeSO_4+(B)KMnO+(C)H_2SO_4 -> (D)Fe_2(SO_4)_3+(E)K_2SO_4+(F)MnSO_4+(G)H_2O` ?

Answer 1

`10FeSO_4+2KMnO+8H_2SO_4 -> 5Fe_2(SO_4)_3+K_2SO_4+2MnSO_4+8H_2O`

Explanation:

Let's eliminate unknowns backwards, starting with `G`.

From the equations for `H`, `Mn`, `K` and `Fe` we find:

`G=C`

`F=B`

`E=1/2 B`

`D=1/2 A`

Substituting these in the remaining equations we get:

`S: A+C=3(1/2 A)+(1/2 B)+B = 3/2A + 3/2B`

`O: 4A+4B+4C = 12 (1/2 A)+4 (1/2 B)+4B+C = 6A + 6B+C`

Subtracting `A` from both sides of the equation for `S` we get:

`C=1/2A+3/2B`

Substitute this in our equation for `O` to get:

`4A+4B+4(1/2A+3/2B) = 6A+6B+(1/2A+3/2B)`

That is:

`6A+10B = 13/2A+15/2B`

Multiplying both sides by `2` that becomes:

`12A+20B = 13A+15B`

Subtract `12A+15B` from both sides to get:

`5B=A`

So if we put `B=2` then we get:

`A=5B=10`

`B=2`

`C=1/2A+3/2B=5+3=8`

`D=1/2A = 5`

`E=1/2B = 1`

`F=B=2`

`G=C=8`

So:

`10FeSO_4+2KMnO+8H_2SO_4 -> 5Fe_2(SO_4)_3+K_2SO_4+2MnSO_4+8H_2O`