Question

calculate the volume of oxygen required for combustion of 0.215dm³ of ethane at 375 k​

Answer 1

Answer:

here is your answer

Explanation:

Volume of Ethane = 0.215 dm3 = 0.215 L

Combustion of ethane – C2H6+2O2--> CO2+2H2O  

We know that , ( Volume of 1 mol gas is 22.4 L )

Amount of oxygen required for the combustion of 22.4L of ethane = (2x22.4)L  = 44.8 L

Amount of oxygen required for the combustion of 0.215L of ethane =(44.8/22.4x0.215 )   = 0.5 L (approx.)

So, 0.5 L Oxygen required for the complete combustion of 0.215 dm3 of methane.