Balancing the burning of butane

Take a look at an equation showing the burning of butane, a hydrocarbon, with excess oxygen available. (This is the reaction that takes place when you light a butane lighter.) The unbalanced reaction is

C₄H_10(g) + O_2(g) → CO_2(g) + H₂O_(g)

Because waiting until the end to balance hydrogen atoms and oxygen atoms is always a good idea, balance the carbon atoms first. You have four carbon atoms on the left and one carbon atom on the right, so add a coefficient of 4 in front of the carbon dioxide:

C₄H_10(g) + O_2(g) → 4CO_2(g) + H₂O_(g)

Balance the hydrogen atoms next. You have ten hydrogen atoms on the left and two hydrogen atoms on the right, so use a coefficient of 5 in front of the water on the right:

C₄H_10(g) + O_2(g) → 4CO_2(g) + 5H₂O_(g)

Now work on balancing the oxygen atoms. You have two oxygen atoms on the left and a total of thirteen oxygen atoms on the right [(4 × 2) + (5 × 1) = 13]. What can you multiply 2 by to equal 13? How about 6.5?

C₄H_10(g) + 6.5O_2(g) → 4CO_2(g) + 5H₂O_(g)

But you’re not done. You want the lowest whole-number ratio of coefficients. Multiply the entire equation by 2 to generate whole numbers:

[C₄H_10(g) + 6.5O_2(g) → 4CO_2(g) + 5H₂O_(g)] × 2

Multiply every coefficient by 2 (don’t touch the subscripts!) to get

2 C₄H_10(g) + 13O_2(g) → 8CO_2(g) + 10 H₂O_(g)

If you check the atom count on both sides of the equation, you find that the equation is balanced, and the coefficients are in the lowest whole-number ratio.

After balancing an equation, make sure that the same number of each atom is on both sides and that the coefficients are in the lowest whole-number ratio.