You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
#"CH"_4"O" + "O"_2 → "CO"_2 + "H"_2"O"#
A method that often works is to balance everything other than #"O"# and #"H"# first, then balance #"O"#, and finally balance #"H"#.
Another useful procedure is to start with what looks like the most complicated formula.
The most complicated formula looks like #"CH"_4"O"#. We put a #1# in front of it to remind ourselves that the coefficient is now fixed.
We start with
#color(red)(1)"CH"_4"O" + "O"_2 → "CO"_2 + "H"_2"O"#
Balance #"C"#:
We have fixed 1 #"C"# atom on the left-hand side, so we need 1 #"C"# atom on the right-hand side. We put a #1# in front of the #"CO"_2#.
#color(red)(1)"CH"_4"O" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"#
Balance #"O"#:
We can't balance #"O"# yet, because we have two formulas that contain #"O"# and lack coefficients. So we balance #"H"# instead.
Balance #"H"#:
We have fixed 4 #"H"# atoms on the left-hand side, so we need 4 #"H"# atoms on the right-hand side. We put an #2# in front of the #"H"_2"O"#.
#color(red)(1)"CH"_4"O" + "O"_2 → color(blue)(1)"CO"_2 + color(green)(2)"H"_2"O"#
Now we balance #"O"#:
We have fixed 4 #"O"# atoms on the right-hand side: 2 from the #"CO"_2# and 2 from the #"H"_2"O"#.
We have fixed 1 #"O"# atom on the left-hand side, so we need 3 more from the #"O"_2#.
Oops. We would need a fraction of an #"O"_2# molecule.
We start over, multiplying every coefficient by 2.
#color(red)(2)"CH"_4"O" + "O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O"#
Now we have fixed 8 #"O"# atoms on the right and 2 on the left. We need 6 more #"O"# atoms on the left. We put a 3 in front of the #"O"_2#.
#color(red)(2)"CH"_4"O" + color(brown)(3)"O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O"#
Every formula now has a fixed coefficient. We should have a balanced equation.
Let’s check:
#mathbf(color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side")#
#color(white)(mmll)"C"color(white)(mmmmmml)2color(white)(mmmmmmmmll)2#
#color(white)(mmll)"H"color(white)(mmmmmml)8color(white)(mmmmmmmmll)8#
#color(white)(mmll)"O"color(white)(mmmmmml)8color(white)(mmmmmmmmll)8#
All atoms balance. The balanced equation is
#"2CH"_4"O" + "3O"_2 → "2CO"_2 + "4H"_2"O"#
