Honourable Ma'am, I think you have done a calculation mistakes in the following : nE0 in the last 2 questions. If it's not a mistake, then please explain me...Hope to reply as early as possible... Waiting for your...
Yes you are right. I have to take stoichiometric coefficient of ethylene oxide as 2. Then nEo will be 50 kmol in the first case when fractional conversion of the limiting reactant is 50%. In the second case when 60 kmol of oxygen is left, extent of reaction is 40, so nEo will be 2 x 40 = 80 kmol.
Honourable Ma'am, I think you have done a calculation mistakes in the following : nE0 in the last 2 questions. If it's not a mistake, then please explain me...Hope to reply as early as possible... Waiting for your...
Yes you are right. I have to take stoichiometric coefficient of ethylene oxide as 2. Then nEo will be 50 kmol in the first case when fractional conversion of the limiting reactant is 50%. In the second case when 60 kmol of oxygen is left, extent of reaction is 40, so nEo will be 2 x 40 = 80 kmol.
Thank you very much Doctor for the great explanation.. I have one Q.. at the end of the video, How did you get the 75 kmol of n02
Keep up the good work madam