# 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?

Jan 31, 2015

Start by looking at the mole ratio between ${C}_{8} {H}_{18}$ and ${O}_{2}$; notice that you need 25 moles of "O_2 for every 2 moles of ${C}_{8} {H}_{18}$. SInce you start with $\text{4 moles}$ of ${C}_{8} {H}_{18}$, the number of ${O}_{2}$ moles you'll need is

${\text{4.00 moles C"_8"H"_18 * ("25 moles O"_2)/("2 moles C"_8"H"_18) = "50.0 moles O}}_{2}$

Now all you need to do is use the ideal gas law equation, $\text{PV" = "nRT}$, to solve for the volume of ${O}_{2}$ needed (don't forget to transform degrees Celsius to Kelvin)

$P V = n R T \implies V = \frac{n R T}{P}$

${V}_{{O}_{2}} = \left(\text{50.0 moles" * "0.082" ("L" * "atm")/("mol" * "K") * ("273.15 + 35")"K")/("0.953 atm}\right)$

${V}_{{O}_{2}} = \text{1325.7 L}$, or ${V}_{{O}_{2}} = \text{1330 L}$ - rounded to three sig figs.