A compound with molecular formula #CH_4O# burns in air to form carbon dioxide and water. How do you write a balanced equation for this reaction?

1 Answer
Jan 1, 2017

Answer:

Warning! Long answer! The balanced equation is
#"2CH"_4"O" + "3O"_2 → "2CO"_2 + "4H"_2"O"#

Explanation:

You follow a systematic procedure to balance the equation.

Start with the unbalanced equation:

#"CH"_4"O" + "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 #"CH"_4"O"#. We put a #1# in front of it to remind ourselves that the coefficient is now fixed.

We start with

#color(red)(1)"CH"_4"O" + "O"_2 → "CO"_2 + "H"_2"O"#

Balance #"C"#:

We have fixed 1 #"C"# atom on the left-hand side, so we need 1 #"C"# atom on the right-hand side. We put a #1# in front of the #"CO"_2#.

#color(red)(1)"CH"_4"O" + "O"_2 → color(blue)(1)"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 4 #"H"# atoms on the left-hand side, so we need 4 #"H"# atoms on the right-hand side. We put an #2# in front of the #"H"_2"O"#.

#color(red)(1)"CH"_4"O" + "O"_2 → color(blue)(1)"CO"_2 + color(green)(2)"H"_2"O"#

Now we balance #"O"#:

We have fixed 4 #"O"# atoms on the right-hand side: 2 from the #"CO"_2# and 2 from the #"H"_2"O"#.

We have fixed 1 #"O"# atom on the left-hand side, so we need 3 more from the #"O"_2#.

Oops. We would need a fraction of an #"O"_2# molecule.

We start over, multiplying every coefficient by 2.

#color(red)(2)"CH"_4"O" + "O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O"#

Now we have fixed 8 #"O"# atoms on the right and 2 on the left. We need 6 more #"O"# atoms on the left. We put a 3 in front of the #"O"_2#.

#color(red)(2)"CH"_4"O" + color(brown)(3)"O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O"#

Every formula now has a fixed coefficient. We should have a balanced equation.

Let’s check:

#mathbf(color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side")#
#color(white)(mmll)"C"color(white)(mmmmmml)2color(white)(mmmmmmmmll)2#
#color(white)(mmll)"H"color(white)(mmmmmml)8color(white)(mmmmmmmmll)8#
#color(white)(mmll)"O"color(white)(mmmmmml)8color(white)(mmmmmmmmll)8#

All atoms balance. The balanced equation is

#"2CH"_4"O" + "3O"_2 → "2CO"_2 + "4H"_2"O"#