Question #43bb5

Notice that the problem provides you with moles of ethanol, #"CH"_3"CH"_2"OH"#, and asks for grams of oxygen gas, #"O"_2#.

This tells you that the first thing to do here is to calculate how many moles of oxygen gas would be needed, then use a conversion factor to take you from moles to grams.

As you can see from the balanced chemical equation

#"CH"_ 3"CH"_ 2"OH"_ ((g)) + color(blue)(3)"O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((g))#

every mole of ethanol that undergoes combustion requires #color(blue)(3)# moles of oxygen gas. This ratio is preserved regardless of the actual number of moles of ethanol, meaning that you will always have three times more moles of oxygen gas than of ethanol taking part in the reaction.

In your case, #2.5# moles of ethanol would require

#2.5 color(red)(cancel(color(black)("moles CH"_3"CH"_2"OH"))) * (color(blue)(3)color(white)(a)"moles O"_2)/(1color(red)(cancel(color(black)("mole CH"_3"CH"_2"OH")))) = "7.5 moles O"_2#

Now, in order to convert between moles and grams, you need to use the molar mass of oxygen gas as a conversion factor.

Oxygen gas has a molar mass of approximately #"32.0 g mol"^(-1)#, which means that every mole of oxygen gas has a mass of #"32.0 g"#.

In your case, #7.5# moles will have a mass of

#7.5 color(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)("240 g O"_2)color(white)(a/a)|)))#

The answer is rounded to two sig figs.