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
Jul 4, 2017

Well we will treat gasoline as octanes......C_8H_18, and we get a volume of

Explanation:

We need (i) a stoichiometric equation........

C_8H_18(l) +25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)

And so we need 25/2 equiv dioxygen gas.......

And thus if 4*mol of octane are combusted we need 25/2xx4=50*mol "dioxygen gas."

And (ii) given V=(nRT)/P=(50*molxx0.0821*(L*atm)/(K*mol)xx308*K)/(0.953*atm)

Approx....1300*L or 1.3*m^3 "dioxygen gas".