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 .......