How do you write the complete equation for #C_10H_20(l) + O_2(g)#?
1 Answer
Explanation:
The first thing to do here is figure out what the products of this reaction will be.
As you can see, you're dealing with the combustion of a hydrocarbon, which is a compound that only contains carbon and hydrogen.
The complete combustion of a hydrocarbon will always produce carbon dioxide,
#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])#
An important thing to realize here is that you have an isolated oxygen molecule on the reactants' side. This means that you can use fractional coefficients as a tool to get the balanced chemical equation.
So, focus on balancing the carbon and hydrogen atoms first.
The thing to remember here is that every atom that is found on the reactants' side must be accounted for on the products' side.
Notice that you have
Multiply this molecule by
#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + "H"_2"O"_text((l])#
Now balance the hydrogen atoms. You know that you have
Once again, multiply the water molecule by
#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#
Finally, focus on balancing the oxygen atoms. Notice that you have
#color(red)(10) xx 2 + color(blue)(10) xx 1 = 30#
atoms of oxygen on the products' side. This means that you must multiply the oxygen molecule by
#"C"_10"H"_text(20(l]) + color(green)(15)"O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#