2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

1 Answer
Jan 31, 2015

Start by looking at the mole ratio between #C_8H_18# and #O_2#; notice that you need 25 moles of #"O_2# for every 2 moles of #C_8H_18#. SInce you start with #"4 moles"# of #C_8H_18#, the number of #O_2# moles you'll need is

#"4.00 moles C"_8"H"_18 * ("25 moles O"_2)/("2 moles C"_8"H"_18) = "50.0 moles O"_2#

Now all you need to do is use the ideal gas law equation, #"PV" = "nRT"#, to solve for the volume of #O_2# needed (don't forget to transform degrees Celsius to Kelvin)

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

#V_(O_2) = ("50.0 moles" * "0.082" ("L" * "atm")/("mol" * "K") * ("273.15 + 35")"K")/("0.953 atm")#

#V_(O_2) = "1325.7 L"#, or #V_(O_2) = "1330 L"# - rounded to three sig figs.