Question

A vessel contains a gaseous mixture of composition by volume 80% H2, and 20% CO. It is desired that the mixture should be made in the proportion 50% H2 and 50% CO by removing some of the mixture and adding some CO. Calculate, per kilomole of mixture, the mass of mixture to be removed and the mass of CO to be added. The pressure and temperature in the vessel remain constant during the process. Take the molar mass of H2 and CO as 2 kg/kmol and 28 kg/kmol, respectively. Hint: Since pressure, temperature, and volume did not change, the amount of substance (number of kilomoles) must remain the same throughout.

Answer 1

Detailed solution has been given below. However, if still any confusion or doubt, feel free to put it in comments section. I'll be happy to help you out with that.

800) Lell.subpose. that..initially, Inixture_con!d:05_*stal Therefore, He presentin mixture = 0.8 kmol LOL 1 kmols. co_present_in mixture : 0.2 kanol of_50% 507 H2 and Now in order to make a mixture 50:2.00, Jel' suppose X kmals_of mixture is removed_and_y_ kmols_of co_is_oddech So, after removing oc kmol of mixtaires. He left in mixture - (6•8-0.8x) kmal of co left ip_mixture =(0.2-0.21) kmol Moles of amount Now after_adding y Imol of C029 in mixture Moles Of Hz (final) - 0:8-0.8% = 0.8(1-7) kmod Moles of co (final) - 0-2-0.238 ty = 0.2(1-x) ty kmal Now total moles in mixture must remain same 507 [xc=y] -0 {no' of moles odded enoot moles remaca) Also , since mixture composition is 50 %. -50% now + Moles.of. Hz (Eina) - 1 = 0·8(1-X Moles of CO 0:2(1-x)+y putting_y=2 in ecyn @ it becomes 0.8-0.8% = 0.2-0.20 tu 0.6. 1.6.X X = 0.6/1:62 = 0.375 kmol -