# What mass of carbon dioxide will be produced by "15.6 g C" if oxygen gas is in excess? "C + O"_2"rarr"CO"_2"

Mar 15, 2016

$\text{15.6 g C}$ will produce $\text{5.72 g CO"_2}$ when oxygen gas is present in excess.

#### Explanation:

Balanced Equation

$\text{C"+"O"_2}$$\rightarrow$$\text{CO"_2}$

Since there is oxygen gas left over, it is the reactant in excess. This means that carbon is the limiting reagent, so we will use the given mass of carbon and stoichiometry to determine how much carbon dioxide can be produced.

We need the molar masses of $\text{C}$ and $\text{CO"_2}$, as well as the mole ratio between $\text{C}$ and $\text{CO"_2}$.

Molar Masses
$\text{C} :$$\text{12.011 g/mol}$ (periodic table)
$\text{CO"_2} :$$\text{44.0095 g/mol}$
https://pubchem.ncbi.nlm.nih.gov/compound/280section=Top

Mole Ratio
$\text{1 mol C} :$$\text{1 mol CO"_2}$

Solution
We will need to make the following conversions using stoichiometry:

$\textcolor{red}{\text{mass C}}$$\rightarrow$$\textcolor{b l u e}{\text{moles C}}$$\rightarrow$$\textcolor{g r e e n}{\text{moles CO"_2}}$$\rightarrow$$\textcolor{p u r p \le}{\text{mass CO"_2}}$

color(red)(15.6"C")xx(color(blue)(1"mol C"))/(color(blue)(12.011"g C"))xx(color(green)(1"mol CO"_2))/(color(green)(1"mol C"))xx(color(purple)(44.0095"g CO"_2))/(color(purple)("mol CO"_2))="5.72 g CO"_2"# rounded to three significant figures

Note: Some people will leave out the mole ratio since it is $1 : 1$, but I think it is important to show how we get from moles of carbon to moles of carbon dioxide.