Calculate the oxidation number of sulfur in na2s2o5. +5

We know that the oxidation number of sodium Na is +1, and oxygen O is -2, therefore:

2 (+1) + 2 x + 5 (-2) = 0

2 + 2x – 10 = 0

2x = 8

x = 4

The oxidation number is +4

The oxidation number of sulfur in Na2S2O5 is +4

HOW TO CALCULATE OXIDATION NUMBER:

Naturally;

Na2S2O5

+1 (2) + x(2) - 2(5) = 0

2 + 2x - 10 = 0

2 + 2x = 10

2x = 10 - 2

2x = 8

x = 4.

Therefore, the oxidation number of sulfur in Na2S2O5 is +4

Learn more: link

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

the answer is c i believe