In a combustion reaction, #"C"# and #"H"# are converted to #"CO"_2# and #"H"_2"O"#.
Start by writing the unbalanced equation.
#"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#
Put a #1# in front of the most complicated formula, #"C"_4"H"_10#.
#color(red)(1)"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#
Step 1. Balance #"C"#.
Put a #4# in front of #"CO"_2#.
#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + "H"_2"O"#
Step 2. Balance #"H"#.
Put a #5# in front of #"H"_2"O"#.
#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + color(green)(5)"H"_2"O"#
Step 3. Balance #"O"#.
We have 13 #"O"# atoms on the right and 2 #"O"# atoms on the left.
We can't balance #"O"# without using fractions.
We start over, multiplying each coefficient by #2#.
#color(red)(2)"C"_4"H"_10 + "O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#
Balance #"O"# atoms by putting #13# in front of #"O"_2#.
# color(red)(2)"C"_4"H"_10 + color(blue)(13)"O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#
Every formula now has a coefficient.
Step 4. Check that atoms are balanced.
#"Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side"#
#color(white)(m)"C" color(white)(mmmmmll)8color(white)(mmmmmmml)8#
#color(white)(m)"H"color(white)(mmmmm)20color(white)(mmmmmmm)20#
#color(white)(m)"O"color(white)(mmmmm)26color(white)(mmmmmmm)26#
All atoms are balanced.
The balanced equation is
#2"C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O"#
