# Question 2e55e

Jul 20, 2017

$18$

#### 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.

${\text{C"_ 8"H"_ (18(g)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O}}_{\left(g\right)}$

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!

${\text{C"_ 8"H"_ (18(g)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + "H"_ 2"O}}_{\left(g\right)}$

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.

${\text{C"_ 8"H"_ (18(g)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O}}_{\left(g\right)}$

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 $\frac{25}{2}$ in order to get $25$ oxygen atoms.

This will get you

${\text{C"_ 8"H"_ (18(g)) + 25/2"O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O}}_{\left(g\right)}$

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

$2 {\text{C"_ 8"H"_ (18(g)) + (2 * 25/2)"O"_ (2(g)) -> (2 * 8)"CO"_ (2(g)) + (2 * 9)"H"_ 2"O}}_{\left(g\right)}$

You will end up with

$2 {\text{C"_ 8"H"_ (18(g)) + 25"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O}}_{\left(g\right)}$