The unbalanced chemical equation is
#"CH"_3"CH"_2"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 often is to start with what looks like the most complicated formula.
The most complicated formula looks like #"CH"_3"CH"_2"OH"#. We put a 1 in front of it to remind ourselves that the number is now fixed.
#color(red)(1)"CH"_3"CH"_2"OH" + "O"_2 → "CO"_2 + "H"_2"O"#
Balance #"C"#:
We have #"2 C"# on the left, so we need #"2 C"# on the right. We put a 2 in front of the #"CO"_2#.
#color(red)(1)"CH"_3"CH"_2"OH" + "O"_2 → color(orange)(2)"CO"_2 + "H"_2"O"#
Balance #"O"#:
We can't balance #"O"# because there are two compounds containing #"O"# that have no coefficients.
Balance #"H"#:
We have fixed #"6 H"# on the left. We need #"6 H"# on the right. Put a 3 in front of #"H"_2"O"#.
#color(red)(1)"CH"_3"CH"_2"OH" + "O"_2 → color(orange)(2)"CO"_2 + color(blue)(3)"H"_2"O"#
Balance #"O"#
We have fixed #"7 O"# atoms on the right and one on the left. We need six more #"O"# atoms on the left. Put a 3 in front of #"O"_2#.
#color(red)(1)"CH"_3"CH"_2"OH" + color(purple)(3)"O"_2 → color(orange)(2)"CO"_2 + color(blue)(3)"H"_2"O"#
Every formula now has a coefficient. We should have a balanced equation.
Let's check.
#bb("Atom" color(white)(m)"lhs"color(white)(m)"rhs")#
#color(white)(m)"C"color(white)(mmm)2color(white)(mml)2#
#color(white)(m)"H"color(white)(mmm)6color(white)(mml)6#
#color(white)(m)"O"color(white)(mmm)7color(white)(mml)7#
All atoms balance. The balanced equation is
#"CH"_3"CH"_2"OH" + "3O"_2 → "2CO"_2 + "3H"_2"O"#
