Question

How do you write the complete equation for `C_10H_20(l) + O_2(g)`?

Answer 1

`"C"_10"H"_text(20(l]) + 15"O"_text(2(g]) -> 10"CO"_text(2(g]) + 10"H"_2"O"_text((l])`

Explanation:

The first thing to do here is figure out what the products of this reaction will be.

As you can see, you're dealing with the combustion of a hydrocarbon, which is a compound that only contains carbon and hydrogen.

The complete combustion of a hydrocarbon will always produce carbon dioxide, `"CO"_2`, and water, `"H"_2"O"`. This means that the unbalanced chemical equation for this presumed complete combustion will look like this

`"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])`

An important thing to realize here is that you have an isolated oxygen molecule on the reactants' side. This means that you can use fractional coefficients as a tool to get the balanced chemical equation.

So, focus on balancing the carbon and hydrogen atoms first.

The thing to remember here is that every atom that is found on the reactants' side must be accounted for on the products' side.

Notice that you have `10` carbon atoms on the reactants' side, but only `1` on the products' side, as part of the carbon dioxide molecule.

Multiply this molecule by `10` to balance the carbon atoms

`"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + "H"_2"O"_text((l])`

Now balance the hydrogen atoms. You know that you have `20` hydrogen atoms on the reactants' side, but only `2` on the products' side, as part of the water molecule.

Once again, multiply the water molecule by `10` to balance the carbon atoms

`"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])`

Finally, focus on balancing the oxygen atoms. Notice that you have `2` oxygen atoms on the reactants' side as part of the oxygen molecule, but a total of

`color(red)(10) xx 2 + color(blue)(10) xx 1 = 30`

atoms of oxygen on the products' side. This means that you must multiply the oxygen molecule by `15` to get the balanced chemical equation for this reaction

`"C"_10"H"_text(20(l]) + color(green)(15)"O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])`