Question #e3578

1 Answer
Aug 17, 2015

Answer:

You'd need 1300 L.

Explanation:

Start by having a look at the balance d chemica lequation for this combustion reaction

#color(red)(2)C_8H_(18(g)) + color(blue)(25)O_(2(g)) -> 16CO_(2(g)) + 18H_2O_((l))#

Notice that you have a #color(red)(2):color(blue)(25)# mole ratio between octane and oxygen gas. This means that for every mole of octane that you want to burn, you need #25/2# moles of oxygen.

You know that you want to burn 4 moles of octane, which means that you're going to need

#4color(blue)(cancel(color(black)("moles octane"))) * (color(blue)(25)" moles O"""_2)/(color(red)(2)color(blue)(cancel(color(black)("moles octane")))) = "50 moles O"""_2#

Now all you have to do is use the ideal gas law equation to determine what volume of oxygen would contain this many moles at #"0.953 atm"# and #35^@"C"#.

#PV = nRT implies V = (nRT)/P#

#V_(O_2) = (50color(blue)(cancel(color(black)("moles"))) * 0.082(color(blue)(cancel(color(black)("atm"))) * "L")/(color(blue)(cancel(color(black)("mole"))) * color(blue)(cancel(color(black)("K")))) * (273.15 + 35)color(blue)(cancel(color(black)("K"))))/(0.953color(blue)(cancel(color(black)("atm"))))#

#V_(O_2) = "1325.7 L O"""_2#

Now, I'll leave the answer rounded to two sig figs, even though you only gave one sig fig for the moles of octane

#V_(O_2) = color(green)("1300 L")#