# Question 25d7e

Jul 28, 2017

$724$ ${\text{g CO}}_{2}$

#### Explanation:

We're asked to find the mass (in $\text{g}$) of ${\text{CO}}_{2}$ that forms when $235$ ${\text{g C"_8"H}}_{18}$ are burned in air.

To do this, we'll first write the balanced chemical equation for this reaction. This is a combustion reaction, so it follows the general form

$2 \text{C"_8"H"_18 (l) + 25"O"_2(g) rarr 16"CO"_2(g) + 18"H"_2"O} \left(g\right)$

We'll first convert the given mass of octane to moles, using its molar mass ($114.23$ $\text{g/mol}$):

235cancel("g C"_8"H"_18)((1color(white)(l)"mol C"_8"H"_18)/(114.23cancel("g C"_8"H"_18))) = color(red)(2.06 color(red)("mol C"_8"H"_18

Now, we'll use the coefficients of the chemical equation to find the relative number of moles of ${\text{CO}}_{2}$ that form:

color(red)(2.06)cancel(color(red)("mol C"_8"H"_18))((16color(white)(l)"mol CO"_2)/(2cancel("mol C"_8"H"_18))) = color(green)(16.5 color(green)("mol CO"_2

Finally, we can use the molar mass of carbon dioxide ($44.01$ $\text{g/mol}$) to find the number of grams of ${\text{CO}}_{2}$ that can form:

color(green)(16.5)cancel(color(green)("mol CO"_2))((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(724 color(blue)("g CO"_2#