Question

A gasoline mixture containing 60.0% by weight octane (C8H18) and 40.0% by weight hexane (C6H14) has a density of 0.660 g mL–1. Calculate the number of cubic meters of air at STP drawn through the carburetor when one imperial gallon of this gasoline is combusted in an automobile engine, assuming complete combustion of the fuel to carbon dioxide and water vapor. Air contains 21.0% by volume of oxygen. One pound (lb) = 453.59 g; one imperial gallon (IG) of water weighs 10.0 lb; liquid water has a density of 1.00 g mL–1

Answer 1

Mass of 1 imperial gallon of water= 10 lb= 10*0.4535kg = 4.535 kg= 4535 gm

Volume of water= 4535 g/ density (1g/ml)= 4535 ml

Mass of same volume of gasoline =4535*0.66 = 2393 gms

Weight of octane= 40% mass of octane = 2393*60/100 = 1435.8 gms

Molecular weight of octane (C8H18)= 8*12+18=114 and hexane (C6H14)=6*12+14= 86

Moles of octane = mass/ Molecular weight= 1435.8/ 114=12.59 mole

Moles of hexane= 2393*40/(100*86) = 11.13

For combustion of Octane (C8H18), the reaction is

C8H18+ 12.5 O2 --> 8CO2 + 9 H2O

1 mole of C8H18 required 12.5 moles of oxygen, i.e 12.5/0.21 moles of air ( air contains 21% by volume of oxygen) =59.52 moles of air

12.59 moles required =12.59*59.52 moles of air =749.36 moles of air

For combustion of Hexane, the reaction is

C614+ 9.5O2 --> 6CO2 + 7H2O

1 mole of hexane requires 9.5/0.21 moles of air =45.24 g moles of air

11.13 moles of hexane required = 45.24*11.13=503.5 g moles of air

Total moles of air to be supplied = 749.36 ( for combustion of octane)+ 503.5 ( for combustion of hexane)= 1253 gmoles of air =12.53 kg moles of air

1 kg mole of any gas at STP occupies 22.4 m3

12.53 kg moles occupy 12.53*22.4=280.672 m3