IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

joyfrancis

A mixture of ethane and ethene occupies 40L at 1.00atm and at 40K. The mixture reacts completely with 130g O₂ gas to produce CO₂ and H₂O . Assuming ideal behaviour , calculte mole fractions of ethane and ethene in the mixture.

neelesh.sahay

C2H6 + 7/2 O2 --------=> 2CO2 + 3 H2O

SUPPOSE WE HAVE N1 MOLE ETHANE IN ORIGINAL MIXTURE SO OXYGEN REQUIRED WOULD BE 7/2 N1

C2H4 + 3 O2 --------=> 2CO2 + 2 H2O

WE HAVE N2 MOLE ETHENE IN ORIGINAL MIXTURE SO O2 REQUIRED WOULD BE = 3 N2

TOTAL O2 = 7/2 N1 + 3 N2 = 130/32 ........EQUATION 1

USING PV= nRT WE CAN CALCULATE NO. OF MOLES OF ETHANE + ETHENE

SO N1 + N2 = 40/40R =1/R WHERE IS GAS CONSTANT IN ATM-LIT /MOL-K

3 N1 + 3 N2 = 3/R .....EQUATION 2

SOLVE THE 2 EQUATIONS U WILL GET N1 AND N2 .......

joyfrancis

ARREY, i made a stupid mistake in balancing!..lol..thnx anyway..rated u

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