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Grade: 11
        
the vapour density of a hydrocarbon is 30.If on complete combustion of hydrocarbon,8.8 g CO2 and 5.4 g H2O are obtained then molecular formula of hydrocarbon is                  1)C3H8 2)C4H12 3)C2H6 4)C2H4
one year ago

Answers : (1)

Shweta Ahire
44 Points
							
Reaction = CxHy + (x + y/4)O2 = xCO2 + y/2 H2O now 8.8 gm CO2 means 8.8/44 moles=0.2 moles 
5.4 gm H2O means 5.4/18 moles = 0.3 moles 
For 1 mole hydrocarbon there is x moles of CO2 so for 0.2 moles of CO2 = 0.2/x moles of hydrocarbon
 
For 1 mole  hydrocarbon there is y/2 mole of H2O so for 0.3 moles of H2O = 0.6/y moles of hydrocarbon 
So now 0.2/x = 0.6/y    it means y=3x..............eq . 1
And molecular weight = 2 × vapour density 
So 12x+ y = 2 × 30 
12x + y = 60..............eq 2
By solving eq 1 and 2 we get x=4  and  y = 12 
So hydrocarbon is C4H12
one year ago
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