How do you balance the complete combustion given by #"C"_4"H"_10 + "O"_2 -> "CO"_2 + "H"_2"O"#? All phases are gases here.

1 Answer
Sep 24, 2017

#color(red)(2)"C"_4"H"_10(g) + color(red)(13)"O"_2(g) -> color(red)(8)"CO"_2(g) + color(red)(10)"H"_2"O"(g)#

What is the state and identity of oxygen gas in nature? Which six other elements are found in this kind of bonding pattern?


A nice trick is that I double the mols of the hydrocarbon, and balance it like that. I therefore avoid having an odd number of oxygens on the left and even number on the right.

The first step is then:

#color(red)(2)"C"_4"H"_10(g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O"(g)#

The hydrocarbon is the only source of carbon in #"CO"_2#. With #8# carbons on the left, I multiply the #"CO"_2# on the right by #8#:

#color(red)(2)"C"_4"H"_10(g) + "O"_2(g) -> color(red)(8)"CO"_2(g) + "H"_2"O"(g)#

The hydrocarbon is the only source of hydrogen in #"H"_2"O"#. With #20# hydrogens on the left, I multiply the water by #10#:

#color(red)(2)"C"_4"H"_10(g) + "O"_2(g) -> color(red)(8)"CO"_2(g) + color(red)(10)"H"_2"O"(g)#

And now the oxygens are easy to balance because you can only affect the oxygen atoms now.

#color(red)(2)"C"_4"H"_10(g) + color(red)(13)"O"_2(g) -> color(red)(8)"CO"_2(g) + color(red)(10)"H"_2"O"(g)#