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Date: 24-1-2017
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Date: 13-4-2019
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Date: 15-2-2019
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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
C4H10(g) + O2(g) → CO2(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:
C4H10(g) + O2(g) → 4CO2(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:
C4H10(g) + O2(g) → 4CO2(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?
C4H10(g) + 6.5O2(g) → 4CO2(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:
[C4H10(g) + 6.5O2(g) → 4CO2(g) + 5H2O(g)] × 2
Multiply every coefficient by 2 (don’t touch the subscripts!) to get
2 C4H10(g) + 13O2(g) → 8CO2(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.
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