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

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#

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...