Start with the unbalanced equation:
#"C"_8"H"_18 + "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 #"C"_8"H"_18#. We put a 1 in front of it to remind ourselves that the number is now fixed.
#color(red)(1)"C"_8"H"_18 + "O"_2 → "CO"_2 + "H"_2"O"#
Balance #"C"#:
We have fixed #"8C"# on the left, so we need #"8 C"# on the right. Put an 8 in front of the #"CO"_2#.
#color(red)(1)"C"_8"H"_18 + "O"_2 → color(blue)(8)"CO"_2 + "H"_2"O"#
Balance #"O"#:
We can't balance #"O"# because we have two formulas containing #"O"# that have no coefficients.
Balance #"H"#:
We have fixed #"18 H"# on the left, so we need #"18 H"# on the right. Put a 9 in front of #"H"_2"O"#.
#color(red)(1)"C"_8"H"_18 + "O"_2 → color(blue)(8)"CO"_2 + color(orange)(9)"H"_2"O"#
Balance #"O"#:
We have fixed #"25 O"# on the right, so we need 25 #"O"# on the left.
Oops! That means we would need #"12.5 O"_2# on the left.
My instructor never let me use fractions when balancing equations.
The solution: Multiply every coefficient by 2.
#color(red)(2)"C"_8"H"_18 + "O"_2 → color(blue)(16)"CO"_2 + color(orange)(18)"H"_2"O"#
Now, we have #"50 O"# on the right, so we can use #"25 O"_2# on the left.
#color(red)(2)"C"_8"H"_18 + color(purple)(25)"O"_2 → color(blue)(16)"CO"_2 + color(orange)(18)"H"_2"O"#
Every formula now has a fixed coefficient. We should have a balanced equation.
Let’s check:
#bb("Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side")#
#color(white)(m)"C"color(white)(mmmmmm)16color(white)(mmmmmmmm)16#
#color(white)(m)"H"color(white)(mmmmmm)36color(white)(mmmmmmmm)36#
#color(white)(m)"O"color(white)(mmmmmm)50color(white)(mmmmmmmm)50#
All atoms balance! The balanced equation is
#2"C"_8"H"_18 +25"O"_2 → 16"CO"_2 + 18"H"_2"O"#
