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
#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
#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
#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)#