Question #09624

Start by writing the balanced chemical equation that describes this combustion reaction

#color(blue)(2)"C"_ 8"H"_ (18(l)) + color(red)(25)"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O"_ ((g))#

Notice that the reaction consumes #color(red)(25)# moles of oxygen gas for every #color(blue)(2)# moles of octane that take part in the reaction.

In other words, regardless of how many moles of octane take part in the reaction, you will always have

#"moles of octane"/"moles of oxygen gas" = color(blue)(2)/color(red)(25)#

To find the mass of oxygen gas needed for the reaction of #"2.27 g"# of octane, convert the mass of octane to moles by using the compound's molar mass

#2.27 color(red)(cancel(color(black)("g"))) * ("1 mole C"_8"H"_18)/(114.23 color(red)(cancel(color(black)("g")))) = "0.01987 moles C"_8"H"_18#

Now use the aforementioned mole ratio to find the number of moles of oxygen gas needed for the reaction

#0.01987 color(red)(cancel(color(black)("moles C"_8"H"_18))) * (color(red)(25)color(white)(a)"moles O"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles C"_8"H"_18)))) = "0.2484 moles O"_2#

To convert the number of moles to grams, use the molar mass of oxygen gas

#0.2484color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("7.95 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.