Question #2e55e

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))#