Im having trouble balancing this redox rection in acidic solution. I don't know how to balance reactions in which one reactant splits up into two products.

C2H5OH(aq) + I3-(aq) --> I-(aq) + CO2(g) + CHO2-(aq) + H+(aq)

I know I split it into
C2H5OH --> CO2 + CHO2 + H+

and I know the oxidations #'s: for the C's:
C2H5OH C=2-
CO2 C=4+
CHO2- C=2+

But, I don't know where to go from there!

One thing you can do here is to combine the loss of electrons from the C2H5OH. If we rewrite that as

C2H6O ==> CO2 + CH2O^-
Each C on the left is -2. C in CO2 is +4 and C in CH2O^- is -1 so we can combine them this way.
C2H6O ==> CO2 + CH2O^-
1 C on left to CO2 is -2 to +4 which is a loss of 6 electrons.
1 C on the left to CH2O^- is -2 to -1 or a loss of 1 electron. Combining them, we have a loss of 6e + 1e = 7e lost for the duo. So the half equation would look like this.
C2H6O ==> CO2 + CH2O^- + 7e
Probably you can finish but we need to add 8H^+ to the right to balance the charge as in this
C2H6O ==> CO2 + CH2O^- + 7e + 8H^+ and that makes us add H2O to the left (we have 10 H on the right so we need 10 on the left).
2H2O + C2H6O ==> CO2 + CH2O^- + 7e + 8H^+.
Now we check it.
I see 10 H on the left and right.
I see 3 O on the left and right.
I see 2 C on left and right.
I see 0 charge on left and right.
Balanced.
Now you balance the I3^- ==> 3I^-

The final equation is
4H2O + 2C2H6O + 7I3^- ==> 21I^- + 2CO2 + 2CH2O^- + 16H^+
Check it to make sure it balances:
1. By atom
2. By charge
3. By electron change.

Thanks! I actually eventually got the right answer after about 45 gruelling minutes.

But, mine is different than yours

3H2O + C2H5OH + %I3- ==> 15I- + CO2 + CHO2- + 11H+

To balance a redox reaction in acidic solution, follow these steps:

1. Split the reaction into two half-reactions: oxidation and reduction.

Oxidation half-reaction:
C2H5OH(aq) -> 2CO2(g)

Reduction half-reaction:
I3-(aq) -> I-(aq)

2. Balance the atoms other than hydrogen and oxygen in each half-reaction.

For the oxidation half-reaction:
C2H5OH(aq) -> 2CO2(g)
The carbon atoms are already balanced.
The hydrogen atoms are balanced: 6H on the left and 6H on the right.
The oxygen atoms are not balanced: 2O on the left and 4O on the right. To balance the oxygen atoms, add H2O molecules to the right side of the equation:
C2H5OH(aq) -> 2CO2(g) + 3H2O(l)

For the reduction half-reaction:
I3-(aq) -> I-(aq)
The iodine atoms are not balanced: 1I on the left and 3I on the right. To balance the iodine atoms, add I- ions to the left side of the equation:
I3-(aq) + 6I-(aq) -> 7I-(aq)

3. Balance the charge in each half-reaction by adding electrons (e-).

For the oxidation half-reaction:
C2H5OH(aq) -> 2CO2(g) + 3H2O(l)
The charge is neutral on both sides, so no electrons are needed.

For the reduction half-reaction:
I3-(aq) + 6I-(aq) -> 7I-(aq)
The charge on the left side is 1- and the charge on the right side is 7-. To balance the charge, add 6 electrons (e-) to the left side of the equation:
I3-(aq) + 6I-(aq) + 6e- -> 7I-(aq)

4. Multiply each half-reaction by integers to equalize the number of electrons in both half-reactions (if necessary).

In this case, the number of electrons is already equal in both half-reactions.

5. Combine the half-reactions.

To combine the half-reactions, multiply each half-reaction by the appropriate factor so that the number of electrons cancels out when they are added together.

2(C2H5OH(aq) -> 2CO2(g) + 3H2O(l)) multiplied by 6
6C2H5OH(aq) -> 12CO2(g) + 18H2O(l)

(I3-(aq) + 6I-(aq) + 6e- -> 7I-(aq)) multiplied by 2
2I3-(aq) + 12I-(aq) + 12e- -> 14I-(aq)

Adding the balanced half-reactions together, the electrons cancel out, and you obtain the balanced overall redox reaction:

6C2H5OH(aq) + 2I3-(aq) + 12I-(aq) + 12e- -> 12CO2(g) + 14I-(aq) + 18H2O(l)

Simplifying the equation, you get:

C2H5OH(aq) + I3-(aq) -> 2CO2(g) + 12I-(aq) + 9H2O(l)

Make sure to double-check your work and ensure that the number of each type of atom is balanced on both sides of the equation.