We're asked to find the mass (in "g"g) of "CO"_2CO2 that forms when 235235 "g C"_8"H"_18g C8H18 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"C"_8"H"_18 (l) + 25"O"_2(g) rarr 16"CO"_2(g) + 18"H"_2"O"(g)2C8H18(l)+25O2(g)→16CO2(g)+18H2O(g)
We'll first convert the given mass of octane to moles, using its molar mass (114.23114.23 "g/mol"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 "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 "g/mol") to find the number of grams of "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
