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

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

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Answer 1 · 2018-07-01T16:51:49

$8 N a_{2} S O_{3} + S_{8} \to 8 N a_{2} S_{2} O_{3}$ $8Na_2SO_3 + S_8 → 8Na_2S_2O_3$

Explanation:

It's Actually easy.

First Let's Check The Equation.

$N a_{2} S O_{3} + S_{8} \to N a_{2} S_{2} O_{3}$ $Na_2SO_3 + S_8 -> Na_2S_2O_3$

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 $N a_{2} S O_{3}$ $Na_2SO_3$ is $+ 4$ $+4$ , one of $S_{8}$ $S_8$ in $0$ $0$ and one of $N a_{2} S_{2} O_{3}$ $Na_2S_2O_3$ in $+ 2$ $+2$ .

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

The key changes are:

$S_{8} \to 8 S^{+ 2} + 16 e^{-}$ $S_8 → 8S^(+2)+16e^(-)$ and
$S^{+ 4} + 2 e^{-} \to S^{+ 2}$ $S^(+4)+2e^(-)→ S^(+2)$

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

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

$8 N a_{2} S O_{3} + S_{8} \to 8 N a_{2} S_{2} O_{3}$ $8Na_2SO_3 + S_8 → 8Na_2S_2O_3$

Total Balance:

$L e f t R i g h t$ $" " Left" "Right$
$N a 1616$ $Na" 16" " "16$
$S 1616$ $S" " 16 " "16$
$O 2424$ $O" 24" " "24$

Answer 2 · 2018-07-01T17:51:37

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

Explanation:

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

And so reduction....

$2 S O_{3}^{2 -} + 6 H^{+} + 4 e^{-} \to S_{2} O_{3}^{2 -} + 3 H_{2} O$ $2SO_3^(2-)+6H^+ +4e^(-) rarr S_2O_3^(2-)+3H_2O$

And oxidation...

$\frac{1}{4} S_{8} + 3 H_{2} O \to S_{2} O_{3}^{2 -} + 6 H^{+} + 4 e^{-}$ $1/4S_8 +3H_2Orarr S_2O_3^(2-)+6H^+ +4e^(-)$

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

$2 S O_{3}^{2 -} + 6 H^{+} + 4 e^{-} + \frac{1}{4} S_{8} + 3 H_{2} O \to S_{2} O_{3}^{2 -} + 3 H_{2} O + S_{2} O_{3}^{2 -} + 6 H^{+} + 4 e^{-}$ $2SO_3^(2-)+6H^+ +4e^(-) +1/4S_8 +3H_2Orarr S_2O_3^(2-)+3H_2O+S_2O_3^(2-)+6H^+ +4e^(-)$

And we cancel common reagents...

$2 S O_{3}^{2 -} + \frac{1}{4} S_{8} \to 2 S_{2} O_{3}^{2 -}$ $2SO_3^(2-) +1/4S_8 rarr 2S_2O_3^(2-)$

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

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

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