Question

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"`?

Answer 1

`"1.02 mol O"_2(g)`

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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)`.