# 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"?

Dec 19, 2016

${\text{1.02 mol O}}_{2} \left(g\right)$

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

$\text{C"("s") + "O"_2(g) -> "CO} \left(g\right)$

Note that ${\text{O}}_{2}$ 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 $\text{CO}$ molecules, so that we maintain the structure of the compounds and elements in the reaction:

$\implies \text{C"("s") + "O"_2(g) -> color(red)(2)"CO} \left(g\right)$

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:

$\implies \textcolor{b l u e}{\textcolor{red}{2} \text{C"("s") + "O"_2(g) -> color(red)(2)"CO} \left(g\right)}$

Since in the actual reaction in the scenario, we have $\text{2.04 mols}$ of "C"("s"), we should expect to have a little over $\text{1.00 mol}$ of ${\text{O}}_{2} \left(g\right)$ reacting.

The actual math gives:

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

$= \textcolor{b l u e}{{\text{1.02 mols O}}_{2} \left(g\right)}$

Indeed, a little over $\text{1.00 mol}$ of ${\text{O}}_{2} \left(g\right)$.