Question

What volume of CO2 is produced from combustion of 1.00 gal of gasoline (2.65 kg) during a hot day in August (35 °C) at sea level (1.00 atm barometric pressure)? What volume of O2 from the air was consumed? Assume gasoline is pure isooctane (C8H18).

Answer 1

C8H8 + 10 O2 ..........>8 CO2 + 4 H2O

no.of moles of gasoline = wt/mol.wt = 2.65*10^3/114 = 23.24 moles

As per the balanced eq.

1 mole of Gasoline can give 8 moles of CO2

23.24 moles of gasoline .......................?

    = 23.24*8/1 = 186 moles of CO2

As per the PV = nRT

V = 186*0.0821*308/1 = 4703.3 lt

..................................................................

1mole of gasoline requires 10 moles of O2

23.24 moles of gasoline ................?

    = 23.24*10 = 232.4 moles

V = nRT/P

V = 232.4*0.0821*308/1 = 5876.6 lt

5876.6 lt of O2 from the air was consumed