For the reaction $C (s) + O_{2} (g) \to CO (g)$ $"C"(s) + "O"_2(g) -> "CO"(g)$ , how many mols of $O_{2}$ $"O"_2$ would be made from $2.04 mols$ $"2.04 mols"$ of $C$ $"C"$ ?

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For the reaction $C (s) + O_{2} (g) \to CO (g)$ $"C"(s) + "O"_2(g) -> "CO"(g)$ , how many mols of $O_{2}$ $"O"_2$ would be made from $2.04 mols$ $"2.04 mols"$ of $C$ $"C"$ ?

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Answer 1 · 2016-12-19T22:42:30

${1.02 mol O}_{2} (g)$ $"1.02 mol O"_2(g)$

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

$C (s) + O_{2} (g) \to CO (g)$ $"C"("s") + "O"_2(g) -> "CO"(g)$

Note that $O_{2}$ $"O"_2$ 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 $CO$ $"CO"$ molecules, so that we maintain the structure of the compounds and elements in the reaction:

$\Rightarrow C (s) + O_{2} (g) \to 2 CO (g)$ $=> "C"("s") + "O"_2(g) -> color(red)(2)"CO"(g)$

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:

$\Rightarrow 2 C (s) + O_{2} (g) \to 2 CO (g)$ $=> color(blue)(color(red)(2)"C"("s") + "O"_2(g) -> color(red)(2)"CO"(g))$

Since in the actual reaction in the scenario, we have $2.04 mols$ $"2.04 mols"$ of $C (s)$ $"C"("s")$ , we should expect to have a little over $1.00 mol$ $"1.00 mol"$ of $O_{2} (g)$ $"O"_2(g)$ 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 $"O"_2(g)$ .

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For the reaction $C (s) + O_{2} (g) \to CO (g)$ $"C"(s) + "O"_2(g) -> "CO"(g)$ , how many mols of $O_{2}$ $"O"_2$ would be made from $2.04 mols$ $"2.04 mols"$ of $C$ $"C"$ ?

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