Place the numbers in the correct position in the equation to balance the reaction. Some numbers may be used more than once.

____$$CH_4$$ + ____$$O_2->$$____$$CO_2$$ + ____$$H_2O$$

1, 2, 3, 4

To balance chemical reactions, we must have the same number of atoms of each type of element on the left side of the arrow compared to the right side.

Three rules we must first understand and follow:

1. Subscripts indicate the amount of atoms of that element per molecule. Ex. The 2 in $H_{2}O$ indicates that there are 2 hydrogen atoms per every molecule of water.
2. Coefficients indicate the number of that specific type of molecule. Ex. The 3 in $3H_{2}O$ indicates 3 molecules of water.
3. To find the total number of atoms of that element type, we must multiply the coefficient by the subscript. Ex. In 3 molecules of water, $3H_{2}O$, there are a total of 6 hydrogen (H) atoms (2$·$3 = 6).

Since hydrogen (H) only exists in one molecule on both the left and the right sides of the reaction, it is easier to balance this one first.

If we assume all coefficients are 1 to begin with, there are 4 atoms of Hydrogen on the left side and only 2 on the right side. Let's add a 2 as a coefficient to $H_{2}0$, giving us a total of 4 hydrogen atoms (2$·$2=4) and keep a 1 as a coefficient for $C{H}_4$. We now have:

$1C{H}_4+\_\_O_2\to \_\_C{O}_2+2H_{2}O$

Since we only have one carbon atom on the left side, lets add a 1 in front of $C{O}_2$ to ensure we only have 1 atom of carbon on the right side of our equation as well.

$1C{H}_4+\_\_O_2\to 1C{O}_2+2H_{2}O$

Finally, lets figure out how many oxygen atoms we have on the completed right side of our equation. We have 1$·$2=2 from the first molecule and 2$·$1=2 from the second. We have 4 total, and therefore need 4 total on the left side as well. Since every $O_2$ molecule will give us 2 oxygen atoms, we need a total of 2 molecules in order to get 4. Below is our final equation:

$1C{H}_4+2O_2\to 1C{O}_2+2H_{2}O$