Question

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?

Answer 1

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.