How do you write a balanced equation for the complete combustion of #C_4H_8#?

1 Answer
Sep 9, 2016

Answer:

#2"C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O"#

Explanation:

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"#