Question #598b4

I'll show you how to solve parts (a) and (b) so that you can practice the technique and solve part (c) on your own.

Start with the balanced chemical equation for this combustion reaction

#2"CH"_3"OH"_text((l]) + color(red)(3)"O"_text(2(g]) -> color(blue)(2)"CO"_text(2(g]) + color(purple)(4)"H"_2"O"_text((l])#

Notice the mole ratios that exist between the species that take part in the reaction. Every 2 moles of methanol will need 3 moles of oxygen gas and produce 2 moles of carbon dioxide and 4 moles of water.

These ratios will be true regardless of how many moles of methanol you start with.

So, how many moles of methanol do you get in your #"154 g"# sample? Use methanol's molar mass to figure that out

#154color(red)(cancel(color(black)("g"))) * "1 mole methanol"/(32.042color(red)(cancel(color(black)("g")))) = "4.806 moles methanol"#

So, using the aforementioned mole ratios, you will need

#4.806color(red)(cancel(color(black)("moles CH"_3"OH"))) * (color(red)(3)" moles O"_2)/(2color(red)(cancel(color(black)("moles CH"_3"OH")))) = "7.209 moles O"_2#

of oxygen gas to make sure that all the moles of methanol react. To get the number of grams of oxygen gas, use its molar mass

#7.209color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)("231 g O"_2#

Likewise, the reaction would produce

#4.806color(red)(cancel(color(black)("moles CH"_3"OH"))) * (color(blue)(2)" moles CO"_2)/(2color(red)(cancel(color(black)("moles CH"_3"OH")))) = "4.806 moles CO"_2#

Use carbon dioxide's molar mass to get the number of grams produced

#4.806color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)("212 g CO"_2#