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

Question

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

Wyatt Adkins · Accepted Answer

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

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

Thus, reduction.

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

Additionally, oxidation...

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

To get rid of the electrons, we combine the equations.

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

We also cancel the common reagents.

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

The which, in my opinion, is balanced in terms of mass and charge—as it really needs to be if we are to accurately reflect chemical reality.

Natalie Anton · Answer

#8Na_2SO_3 + S_8 → 8Na_2S_2O_3#

It really isn't that hard.

Let's check the equation first.

#Na_2SO_3 + S_8 -> Na_2S_2O_3#

There are two sulfur atoms visible in L.H.S., but only one in R.H.S.

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 principal modifications are:

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

Both of these equations require that the second be multiplied by 8 in order to be balanced.

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

#8Na_2SO_3 + S_8 → 8Na_2S_2O_3#

Overall Balance:

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

Cooper Anderson · Answer

undefined The balanced equation is: 3Na2SO3 + 4S8 → 2Na2S2O3

How would you balance: Na2SO3+S8 --&gt; Na2S2O3?

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