For the reaction #"C"(s) + "O"_2(g) -> "CO"(g)#, how many mols of #"O"_2# would be made from #"2.04 mols"# of #"C"#?

#"1.02 mol O"_2(g)#

Pure carbon in charcoal can simply be considered #"C"("s")# (soot). Including the phases, we would have:

#"C"("s") + "O"_2(g) -> "CO"(g)#

Note that #"O"_2# exists in nature as a diatomic gas (two identical oxygen atoms bound together).

Carbon monoxide is a covalent compound, so the "mono" is the prefix that indicates that one oxygen is bound to the carbon in this molecule.

As-is, this reaction is unbalanced. Try to keep a tab on the number of atoms on both sides of the reaction; since there are two oxygen atoms on the left, we need two on the right.

It makes physical sense not to add new subscripts, but to double the number of #"CO"# molecules, so that we maintain the structure of the compounds and elements in the reaction:

#=> "C"("s") + "O"_2(g) -> color(red)(2)"CO"(g)#

If you notice, we've balanced the oxygens and unbalanced the carbons. So, double the number of carbon atoms on the reactants side to get:

#=> color(blue)(color(red)(2)"C"("s") + "O"_2(g) -> color(red)(2)"CO"(g))#

Since in the actual reaction in the scenario, we have #"2.04 mols"# of #"C"("s")#, we should expect to have a little over #"1.00 mol"# of #"O"_2(g)# reacting.

The actual math gives:

#2.04 cancel("mols C"("s")) xx ("1 mol O"_2(g))/(2 cancel("mols C"("s")))#

#= color(blue)("1.02 mols O"_2(g))#

Indeed, a little over #"1.00 mol"# of #"O"_2(g)#.