When the following chemical equation is balanced, what is the coefficient of H2O?

CeO2 + KI + HCl —> KCl + CeCl3 + H2O + I2

Answer:
The answer is 4 and the balanced reaction equation is:
2 CeO2 + 2 KI + 8 HCl —> 2 CeCl3 + 2 KCl + 4 H2O + I2

You have a redox reaction going on between Ce and I:
* Ce(4+) + (e-) —> Ce(3+)
* 2I(-) —> I2 + 2(e-)
In order to balance the electrons (as many on either side of the reaction equation), you have to multiply the first equation by 2. That's why you have 2 in front of CeO2 and CeCl3.

So you already have:
2 CeO2 + 2 KI + x HCl —> 2 CeCl3 + y KCl + z H2O + I2
You have 2 K on the left, so you must have 2 on the right as well, thus y=2. Then you can balance the Cl: you have 2*3+2=8 on the right, so x=8 HCl are required on the left. Now that you have 8 H on the left, you must put z=8/2=4 in front of H2O and you end up with:
2 CeO2 + 2 KI + 8 HCl —> 2 CeCl3 + 2 KCl + 4 H2O + I2
Finally, verify that you have the right number of O on both sides, since it's the only atom you haven't really considered si far: 2*2 on the left, 4 on the right, it works...

I hope that it was clear enough and that it helped!
Whenever you are balancing a reaction and you find yourself going around and around and around, double a reactant. Chances are , because things are found in multiple places, you need to double something to make it work

2CeO2 + 2KI + 8 HCl ---> 2KCl + 2CeCl3 + 4H2O + I2

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