Question

How would you balance the following equation: CH3CH2CH3(g)+O2(g) --> CO2(g) + H2O(g)?

Answer 1

We can rewrite this as:

`"C"_3"H"_8(g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O"(g)`

I would initially start with the number of carbons because the number of carbons in `"CO"_2` is even, and the number of oxygens in `"O"_2` is also even.

`"C"_3"H"_8(g) + "O"_2(g) -> 3"CO"_2(g) + "H"_2"O"(g)`

Then, I would realize that there are `8` hydrogens on the left, and I need `4xx2` on the right.

`"C"_3"H"_8(g) + "O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O"(g)`

Lastly, I would balance the oxygen, because it is easy to only affect the number of oxygens. It only complicates things more to assign a number to `"O"_2` first because you'll probably end up changing it again later.

`color(blue)("C"_3"H"_8(g) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O"(g))`

Last confirmation:

`"C"`: `1xx3 = 3xx1`
`"H"`: `1xx8 = 4xx2`
`"O"`: `5xx2 = 3xx2 + 4xx1`

Since this cannot be reduced any further, this is good to go.