Problem on weight fraction
A gas contains 350 ppm of H2S in CO2 at 72°F and 1.53 atm pressure. If the gas is liquified, what is the weight fraction H2S?
Expert
Now we consider 1liter of CO2 at 72deg F {[K] ≡ (72+ 459.67) × 5/9}=295.3722 K 1.53 atm
So No moles n= PV/RT =1.53atm*1liter/(0.08205liter-atm/K-mole*295.3722) =0.063131mole
So wt of CO = 0.063131mole*28g/mole =1.76767g
Now 350 ppm H2S would have a volume 350 microlitre ( µL ) per litre, = 350 ppm =350E-6 litre
So No of moles H2S = n= PV/RT =1.53atm*(350E-6)liter/(0.08205liter-atm/K-mole*295.3722K) = 2.20959E-05 mole
The MW H2S = 34.116g/mole
So wt of H2S = 0.000753823 g
So total weight =1.768423823g
So %wt H2S = (0.0007538236/1.768423823g)*100%=0.04262%
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Calculate the adiabatic flame temperature of acetylenes gas at a pressure of 1 bar under the following conditions. The reactants are initially at 298K. Assume that the acetylene reacts completely to form CO2 and H2O: Discover Q & A Leading Solution Library Avail More Than 1452996 Solved problems, classrooms assignments, textbook's solutions, for quick Downloads No hassle, Instant Access Start Discovering 18,76,764 1952562 Asked 3,689 Active Tutors 1452996 Questions Answered Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!! Submit Assignment
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