Question

Question #2e55e

Answer 1

`18`

Explanation:

I think that by standard form the problem means without any fractional coefficients, so let's start by balancing the equation with fractional coefficients.

`"C"_ 8"H"_ (18(g)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O"_ ((g))`

It's always a good idea to start with the carbon atoms, so multiply the carbon dioxide molecule by `8` in order to get `8` carbon atoms on the reactants' side--notice that you have `8` carbon atoms on the products' side!

`"C"_ 8"H"_ (18(g)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + "H"_ 2"O"_ ((g))`

Next, balance the hydrogen atoms. Since you have `18` on the reactants' side and only `9` on the products' side, multiply the water molecule by `9` in order to balance the hydrogen atoms out.

`"C"_ 8"H"_ (18(g)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O"_ ((g))`

Now focus on the oxygen atoms. You only have `2` on the reactants' side and a total of

`overbrace(8 xx "2 O atoms")^(color(blue)("from 8 molecules of CO"_2)) + overbrace(9 xx "1 O atom")^(color(blue)("from 9 molecules of H"_2"O")) = "25 O atoms"`

Notice that since you're dealing with an oxygen molecule on the reactants' side, which is comprised of `2` oxygen atoms, you can multiply it by `25/2` in order to get `25` oxygen atoms.

This will get you

`"C"_ 8"H"_ (18(g)) + 25/2"O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O"_ ((g))`

Finally, to get this balanced chemical equation to standard form, multiply all the chemical species by `2` `->` this will get rid of the fractional coefficient that's currently in front of the oxygen molecule.

`2"C"_ 8"H"_ (18(g)) + (2 * 25/2)"O"_ (2(g)) -> (2 * 8)"CO"_ (2(g)) + (2 * 9)"H"_ 2"O"_ ((g))`

You will end up with

`2"C"_ 8"H"_ (18(g)) + 25"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O"_ ((g))`