#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?

1 Answer
Jul 2, 2018

Answer:

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#....