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

Jul 1, 2018

8Na_2SO_3 + S_8 → 8Na_2S_2O_3

#### Explanation:

It's Actually easy.

First Let's Check The Equation.

$N {a}_{2} S {O}_{3} + {S}_{8} \to N {a}_{2} {S}_{2} {O}_{3}$

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 $N {a}_{2} S {O}_{3}$ is $+ 4$, one of ${S}_{8}$ in $0$ and one of $N {a}_{2} {S}_{2} {O}_{3}$ in $+ 2$.

The sulphur is being oxidized (${S}^{0} \to {S}^{2 +}$) and it is being reduced (${S}^{+ 4} \to {S}^{2 +}$). 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:

$\text{ " Left" } R i g h t$
$N a \text{ 16" " } 16$
$S \text{ " 16 " } 16$
$O \text{ 24" " } 24$

Jul 1, 2018

What we gots is a $\text{comproportionation reaction...}$

#### Explanation:

Where sulfur, as $S \left(+ I V\right)$ and $S \left(0\right)$ undergoes a redox reaction to give $S \left(V I +\right)$ and $S \left(- I I\right)$...i.e. $S {\left(+ I I\right)}_{\text{average}}$

And so reduction....

$2 S {O}_{3}^{2 -} + 6 {H}^{+} + 4 {e}^{-} \rightarrow {S}_{2} {O}_{3}^{2 -} + 3 {H}_{2} O$

And oxidation...

$\frac{1}{4} {S}_{8} + 3 {H}_{2} O \rightarrow {S}_{2} {O}_{3}^{2 -} + 6 {H}^{+} + 4 {e}^{-}$

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

$2 S {O}_{3}^{2 -} + 6 {H}^{+} + 4 {e}^{-} + \frac{1}{4} {S}_{8} + 3 {H}_{2} O \rightarrow {S}_{2} {O}_{3}^{2 -} + 3 {H}_{2} O + {S}_{2} {O}_{3}^{2 -} + 6 {H}^{+} + 4 {e}^{-}$

And we cancel common reagents...

$2 S {O}_{3}^{2 -} + \frac{1}{4} {S}_{8} \rightarrow 2 {S}_{2} {O}_{3}^{2 -}$

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