Question

`SO_2(aq) + I_2(aq) + 2H_2O(l) -> SO_4^(2-)(aq) + 2I(aq) + 4H^(+)(aq)`. What is the amount, in moles of iodine in 15.00 `cm^3` of a 0.0010 mol/`dm^3` solution?

Answer 1

You have not balanced that redox equation correctly...

Explanation:

`SO_2(aq) + I_2(aq) + 2H_2O(l) rarr SO_4^(2-) + 2I^(-)+4H^+`

Iodine is REDUCED.....

And I am a bit suspicious of the wording of the question....you gots a `15.00*cm^3` volume of `I_2` of a `0.0010*mol*L^-1` solution...

Now since `"concentration"="moles of solute"/"volume of solution"`...we take the product...

`"moles of solute"="concentration"xx"volume of solution..."`

`15.00*cm^3xx10^-3*L*cm^-3xx0.0010*mol*L^-1=1.5xx10^-5*mol`....