Question #1de47

2 Answers
Mar 30, 2017

Answer:

Answer is 317.41 kJ

Explanation:

The combustion equations are:
CH4 + 2O2 --> CO2 + 2H2O
C3H8 + 5O2 --> 3CO2 + 4H2O

So 1 litre of methane will need 2 litre oxygen and 1 litre of propane will need 5 litres of oxygen. Since 16 litres of oxygen is used, 3 litres of methane (which use 6 litre oxygen) and 2 litres of propane (which uses 10 litre oxygen) is in the mixture.

One mole of gas occupies 22.4 litres

So 3 litres of methane --> 3/22.4 moles = 0.1339 moles
2 litres of propane --> 2/22.4 moles = 0.0893 moles

To find heat released simply multiply the number of moles with the heat of combustion given and add.

Heat released = 0.1139 x 890 + 0.0893 x 2220 = 317.41 kJ

Apr 1, 2017

The balanced equations of combustion reactions are :

#CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l),DeltaH_(CH_4(g))=890kJmol^-1color (red)([1])#

#C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(l),DeltaH_(C_3H_8(g))=220kJmol^-1color (red)([2])#

Let here be x L #CH_4# and (5-x)L #C_3H_8# in 5L gas mixture at STP.

So total volume of oxygen consumed as per stoichiometry of the reactions involved will be #2x+5(5-x)=16#

#=>2x+25-5x=16#

#=>3x=9#

#=>x=3L#

So volume of #CH_4(g)=3L#
and volume of #C_3H_8(g)=2L# at STP.

So corresponding number of moles of gases present int 5Lgas mixture can be obtained by dividing the volumes of gases at STP with 22.4L as any gas of 1mol occupies 22.4L at STP

#n_(CH_4(g))=(3L)/(22.4L)=3/22.4mol#

#n_(C_3H_8(g))=(2L)/(22.4L)=2/22.4mol#

So the amount of heat released due to combustion of the 5L gas mixture

#DeltaH_"mixture"=3/22.4DeltaH_(CH_4(g))+2/22.4DeltaH_(C_3H_8(g))#

#=>DeltaH_"mixture"=(3xx890+2xx2220)/22.4kJ~~317kJ#