# 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?

Jul 2, 2018

You have not balanced that redox equation correctly...

#### Explanation:

$S {O}_{2} \left(a q\right) + {I}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right) \rightarrow S {O}_{4}^{2 -} + 2 {I}^{-} + 4 {H}^{+}$

Iodine is REDUCED.....

And I am a bit suspicious of the wording of the question....you gots a $15.00 \cdot c {m}^{3}$ volume of ${I}_{2}$ of a $0.0010 \cdot m o l \cdot {L}^{-} 1$ solution...

Now since $\text{concentration"="moles of solute"/"volume of solution}$...we take the product...

$\text{moles of solute"="concentration"xx"volume of solution...}$

$15.00 \cdot c {m}^{3} \times {10}^{-} 3 \cdot L \cdot c {m}^{-} 3 \times 0.0010 \cdot m o l \cdot {L}^{-} 1 = 1.5 \times {10}^{-} 5 \cdot m o l$....