# What mass of "CO"_2" will be produced when "17.48 g CO" reacts with excess "O"_2"?

Aug 23, 2017

Under the conditions given, ${\text{27.46 g CO}}_{2}$ will be produced.

#### Explanation:

Balanced Equation

"2CO(g) + O"_2("g")$\rightarrow$$\text{2CO"_2("g")}$

First, determine mol $\text{CO}$ in $\text{17.48 g}$ by dividing its given mass by its molar mass $\left(\text{28.010 g/mol}\right)$. Since molar mass is a fraction, g/mol , you can multiply the given mass by the inverse of the molar mass, mol/g .

Next multiply the moles $\text{CO}$ by the mole ratio between $\text{CO}$ and $\text{CO"_2}$ from the balanced equation with ${\text{CO}}_{2}$ in the numerator.

Finally, multiply mol $\text{CO"_2}$ by its molar mass $\left(\text{44.009 g/mol}\right)$.

17.48color(red)cancel(color(black)("g CO"))xx(1color(red)cancel(color(black)("mol CO")))/(28.010color(red)cancel(color(black)("g CO")))xx(2color(red)cancel(color(black)("mol CO"_2)))/(2color(red)cancel(color(black)("mol CO")))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))="27.46 g CO"_2" (rounded to four significant figures)