Question

How would you balance: Na2SO3+S8 --> Na2S2O3?

Answer 1

`8Na_2SO_3 + S_8 → 8Na_2S_2O_3`

Explanation:

It's Actually easy.

First Let's Check The Equation.

`Na_2SO_3 + S_8 -> Na_2S_2O_3`

In L.H.S, We can see two sulphur atoms, but in R.H.S, we can see only one.

According to this equation, it's a redox reaction. Because value of S in `Na_2SO_3` is `+4`, one of `S_8` in `0` and one of `Na_2S_2O_3` in `+2`.

The sulphur is being oxidized (`S^(0) -> S^(2+)`) and it is being reduced (`S^(+4) -> S^(2+)`). If we balance that charge transfer, we balance the element quantities involved.

The key changes are:

`S_8 → 8S^(+2)+16e^(-)` and
`S^(+4)+2e^(-)→ S^(+2)`

According to these equations, second one must be multiplied by 8 for balance.

Put them all together and check the `S` balance:

`8Na_2SO_3 + S_8 → 8Na_2S_2O_3`

Total Balance:

`" " Left" "Right`
`Na" 16" " "16`
`S" " 16 " "16`
`O" 24" " "24`

Answer 2

What we gots is a `"comproportionation reaction..."`

Explanation:

Where sulfur, as `S(+IV)` and `S(0)` undergoes a redox reaction to give `S(VI+)` and `S(-II)`...i.e. `S(+II)_"average"`

And so reduction....

`2SO_3^(2-)+6H^+ +4e^(-) rarr S_2O_3^(2-)+3H_2O`

And oxidation...

`1/4S_8 +3H_2Orarr S_2O_3^(2-)+6H^+ +4e^(-)`

We add the equations together to eliminate the electrons...

`2SO_3^(2-)+6H^+ +4e^(-) +1/4S_8 +3H_2Orarr S_2O_3^(2-)+3H_2O+S_2O_3^(2-)+6H^+ +4e^(-)`

And we cancel common reagents...

`2SO_3^(2-) +1/4S_8 rarr 2S_2O_3^(2-)`

The which, I think, is balanced with respect to mass and charge...as indeed it must be if we reflect chemical reality...