Answer:
4CrO3====> 2 Cr2O3 + 3O2
Explanation:
Before dealing with the Cr, I think you might want to deal to work with the oxygen in Cr2O3
You might want to turn that into an even number. You could put a 2 in front of the Cr2O3, but looking ahead, I can see that will give you a fractional amount for the O2. So I'll use a 4
CrO3 ===> 4 Cr2O3 + O2
That means you have 8 Cr s on the right. Balance that on the left
8CrO3 =====> 4 Cr2O3 + O2
You've balanced The Cr s. So now lets look at the oxygens. There are 8*3 oxygens on the left hand side. The right has 12 + 2. Don't touch the Cr2O3. You are done with them.
You need 12 more oxygens on the right. You have 2. So multiply the Oxygens by 6
8CrO3 ===> 4 Cr2O3 + 6 O2 There's a factor of 2 common. Divide by 2
4CrO3====> 2 Cr2O3 + 3O2
