You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
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#"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#
A method that often works is first to balance everything other than #"O"# and #"H"#, 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"_3"OH"#. We put a 1 in front of it to remind ourselves that the number is now fixed.
We start with
#color(red)(1)"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#
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Balance #"C"#:
We have #"1 C"# on the left, so we need #"1 C"# on the right. We put a 1 in front of the #"CO"_2#.
#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"#
Balance #"H"#:
We can't balance #"O"# because we have two oxygen-containing molecules without coefficients. ∴ Let's balance #"H"# instead.
We have #"4 H"# on the left, so we need #"4 H"# on the right. There are already #"2 H"# atoms on the right. We must put a 2 in front of the #"H"_2"O"#.
#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + color(orange)(2)"H"_2"O"#
Balance #"O"#:
We have fixed #"4 O"# on the right and #"1 O"# on the left. We need #"3 O"# on the left.
Uh, oh! Fractions!
We start over, this time doubling all the coefficients.
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#color(red)(2)"CH"_3"OH" + "O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#
Now we can balance #"O"# by putting a 3 in front of #"O"_2#
#color(red)(2)"CH"_3"OH" + color(purple)(3)"O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#
Every formula now has a coefficient. We should have a balanced equation.
Let's check.
#"Atom" color(white)(m)"lhs"color(white)(m)"rhs"#
#color(white)(m)"C"color(white)(mml)2color(white)(mm)2#
#color(white)(m)"H"color(white)(mml)8color(white)(mm)8#
#color(white)(m)"O"color(white)(mml)8color(white)(mm)8#
All atoms balance. The balanced equation is
#2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"#
