Question

A liquid fuel containing C, H and O, was burned completely using excess air. The Orsat analysis of the combustion gas shows 4.48% CO2, 11.21% O2 and 84.31% N2, And the molal humidity of the product gas is 0.179 mol water vapor per mol of dry gas. Calculate % composition by weight of the liquid fuel.

Answer 1

Solution taking Basis 100 mol of dry Que gas Pary = 100 mo fuel feed thee god combuster. =100 mols f Pwet = ? Podry Dry Basis composition 4.48 % CO2 11.21% O2 fo فی می یی له Air for А mols 84.21%.NZ So Moles of Na in Product 0.8431X100 Mols gas= 84,31 mols. Since N2 is trte inert so (N₂ in air feed) = Wa in Product gas) air contains 79 % NZ 0.79 A = 84.31 mol A = 106.7215 mols y Moles of air supplied. 00448 X 100 mols of CO2 in product 4.48 mols. Moles - Ct O2 CO2 for I mol со, mol C and I mot og required,

Carban balance (corban in feed) (Corben in CO2 → Moles of Bo Corban consumed in the formation of cas fe 4.48 mots Mass og Carbon in feed mic 12 x fc gram c Molar mass gc = 12 g 1 mol ) mc= 12 x u. ug grom 53.76 grom mc = Given Molah humidity 0.179 mol H2o / mot of dry gas & Moles of H₂O Produced = 0.179 x 100 mol H₂O 17.9 mot H2O H₂ + O₂ → H₂O for 1 mot H₂O He and I mola 2 I mole is required, oo for 17.g mot (° 17.g mols H₂O, 17.g mot of H₂ has to be Consumed) 17.9 x 2 gram O H₂ in the feed c Mass of H₂ in feed a мн 35.8 grom

Now 02 balance in CO2 02 in feed fuel 02 air blo Bel ca in fue gas there) >> fo + 0.21 A = 4.48 + 4 x 17.9 + 11.21 moly 2 - fort 0.21 x 106.7215 - 24.64 maly ros 2.228 mols Molar moss of a = 32 =32 gram/mol 6. Mass flere gas in flu 2.228 X 32 of O2 gram mo 71.311 gram. Oo Total moss of liquid fuel = met mit mo = 53.76 + 35.8 t 71.311 gram 3 = 160.87 gram composition of liquid fuel moss boris Mc C x 100 % 53.76 160.87 7. = 33.418 % *100 Ми x 106 % H - 35.8 x 100 % = 22.25 % 160.87 44,328 % mo 2100 % 7.311 x 100 % 160.87 Answer liquid fuel contains 33.418 % C C 22.25 %H 49.3280