We're asked to find the mass (in #"g"#) of #"CO"_2# that forms when #235# #"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"C"_8"H"_18 (l) + 25"O"_2(g) rarr 16"CO"_2(g) + 18"H"_2"O"(g)#
We'll first convert the given mass of octane to moles, using its molar mass (#114.23# #"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#
