MEEBAL 2012 Quiz 6

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Unformatted text preview: on proceeds to comple.on. Hint: MW of CH4 is 16, MW of O2 is 32 Answers (Part I, Part II): a)  50 kmol/h, 8.3 kmol b)  75 kmol/h, 8.3 kmol c) 75 kmol/h, 12.5 kmol d) 50 kmol/h, 12.5 kmol Quiz #6 (Solution) 2 CH4 + 3 O2 à༎ 2 CO + 4 H2O Theoretical amount (kmol) of O2 needed = 3 (kmol O2)/2 (kmol CH4)× 40.0 (kmol CH4/h )= 60 (kmol O2/h ) 25 % excess O2 = 1.25 × 60 (kmol O2/h ) = 75 (kmol O2/h ) kmoles of CH4 = 320 (kg CH4 )× 1.00 (kmol CH4) /16.0 (kg CH4 ) =20 kmol CH4, Moles of oxygen = 400 (kg O2) × 1.00 (kmol O2) /32.0 (kg O2 ) = 12.5 (kmol O2) Required moles of methane = 2 (kmol CH4)/3 (kmol O2) × 12.5 (kmol O2) = 8.3 (kmol CH4), Oxygen is the limiting reactant. kmoles of CO produced = 12.5 (kmol O2) × 2 (kmol CO) / 3 (kmol O2) =8.3 (kmol CO) (b) is the answer...
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This test prep was uploaded on 04/03/2014 for the course CHME 2010 taught by Professor Tessier during the Fall '08 term at Rensselaer Polytechnic Institute.

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