You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
#"C"_7"H"_16 + "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"_7"H"_16#. We put a #1# in front of it to remind ourselves that the coefficient is now fixed.
We start with
#color(red)(1)"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"#
Balance #"C"#:
We have fixed 7 #"C"# atoms on the left-hand side, so we need 7 #"C"# atoms on the right-hand side. We put a #7# in front of the #"CO"_2#.
#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"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 16 #"H"# atoms on the left-hand side, so we need 16 #"H"# atoms on the right-hand side. We put an #8# in front of the #"H"_2"O"#.
#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#
Now we can balance #"O"#:
We have fixed 22 #"O"# atoms on the right-hand side: 14 from the #"CO"_2# and 8 from the #"H"_2"O"#. We put an #11# in front of the #"O"_2#.
#color(red)(1)"C"_7"H"_16 + color(teal)(11)"O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#
Every formula now has a fixed coefficient. We should have a balanced equation.
Let’s check:
#color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side"#
#color(white)(mmll)"C"color(white)(mmmmml)7color(white)(mmmmmmmmll)7#
#color(white)(mmll)"H"color(white)(mmmmll)16color(white)(mmmmmmmm)16#
#color(white)(mmll)"O"color(white)(mmmmll)22color(white)(mmmmmmmm)22#
All atoms balance. The balanced equation is
#"C"_7"H"_16 + 11"O"_2 → 7"CO"_2 + 8"H"_2"O"#
