# Question #e8c0a

Oct 22, 2016

$\text{C"_10"H"_20"O"_2+14"O"_2 rarr 10"CO"_2 + 10"H"_2"O}$

#### Explanation:

The easiest way to do this is to balance the carbon and hydrogen atoms, and then balance the oxygen atoms.

$\text{C"_10"H"_20"O"_2+"O"_2 rarr "CO"_2 + "H"_2"O}$

We have $10$ $\text{C}$ atoms on the LHS, so we must have the same on the right hand side:

$\text{C"_10"H"_20"O"_2+"O"_2 rarr color(red)(10)"CO"_2 + "H"_2"O}$

We have $20$ $\text{H}$ atoms on the LHS, so that will give us $10$ $\text{H"_2"O}$ molecules:

$\text{C"_10"H"_20"O"_2+"O"_2 rarr color(red)(10)"CO"_2 + color(blue)(10)"H"_2"O}$

On the RHS we have $30$ $\text{O}$ atoms, and on the left, $2$. This means we'll need $14$ ${\text{O}}_{2}$ atoms to balance the equation:

$\text{C"_10"H"_20"O"_2+color(green)(14)"O"_2 rarr color(red)(10)"CO"_2 + color(blue)(10)"H"_2"O}$