1. What volume of water vapor is produced by burning500g of gasoline at STP and
ID: 681358 • Letter: 1
Question
1. What volume of water vapor is produced by burning500g of gasoline at STP and a temp of 18 degreeC. 2. Determine the density of propanegasC3H8, in grams per liter at 22 degreeC and748 too?? The answer to 1 should be 944L H2O and tonumber2.should be 1.79g/L Thanks for your help , I cant seem to fiure these out to getthe right answers!!!! 1. What volume of water vapor is produced by burning500g of gasoline at STP and a temp of 18 degreeC. 2. Determine the density of propanegasC3H8, in grams per liter at 22 degreeC and748 too?? The answer to 1 should be 944L H2O and tonumber2.should be 1.79g/L Thanks for your help , I cant seem to fiure these out to getthe right answers!!!!Explanation / Answer
1) assuming Gasoline formula as C8H18 :: Molar mass = 114 weight = 500g => moles of gasoline taken = 4.3859 Now combustion reaction is : C8H18 +12.5O2 ---> 8CO2 +9H2O so 1 mole of gasoline gives 9 moles of H2O => 4.3859moles of gasoline gives 9*4.3859=39.473moles ofH2O and since PV=nRT [at STP]: T= 298K n= 39.473 P= 1atm => V=39.473*0.0821*298/1 = 965.74 litres [[at T= 18 degrees= 291 K]] V= 39.473*0.0821*291/1 =943.05tres so at STP volume of water liberated = 965.74 litres,and at T=18 degrees Volume of H2O liberated is is943.05 litres -------------------------------------------------------------------------------------------------------- 2) Given T=22degrees = 295K, P=748torr=748/760 atm since PV=nRT => n=w/mol.wt, mol.wt of C3H8 is 44 so n= w/44 => w/44 =PV/RT => w/V =44P/RT = 44(748/760) /(0.082*295) =1.788 ~= 1.79 g/l so density of propane is 1.79 g/l
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