posted by Ellen Cameron on .
a mixture of 10cm^3 of C5H12 and 0.084g of C12H24 was passed with an excess of oxygen. after cooling to room temperature, the reaidual gas was passed through aqueous potassium hydroxide. what volume was absorbed by the alkali
C5H12 + 8O2 ==> 5CO2 + 6H2O
C12H24 + 18O2 ==> 12CO2 + 12H2O
volume CO2 from pentane = 10cc x (5 mols CO2/1 mol C5H12) = 50 cc CO2.
mols dodecene = 0.084/molar mass = approx 5E-4. You should confirm ALL of these calculations since I've estimated here and there.
mols CO2 from dodecene = 5E-4 x (12 mols CO2/1 mol C12H24) = approx 6E-3.
volume CO2 from dodecene = 6E-3 x L/mol = ?
Total CO2 = L from pentane + L from dedecene.
NOTE. L/mol of a gas at room is 22.4 L at STP x (298/273) = ?