Question

1. One mole of propane (C3H8) is burned with 47.6 mol of air to produce a mixture of CO2, H20, O2 and N2. Assume that the air consists only of O2 and N2 and that there are 3.76 kmol of N2 for each kmol of Oz (i.e., the air is 21% O2 and 79% N2 by volume). a) Determine the number of moles of each of the product species in the product mixture. b) Determine the mole fractions of each of the product species in the product mixture. c) Determine the mass fractions of each of the product species in the product mixture.

Answer 1

solution • Balanced chemical reaction C3 Hq +476 (0₂ + ? 3CO2 + 4H₂O + 42.60₂ +178.976N, - CO2 Ho O N₂ > 3 nole 4 male → 42.6 mole → 178.976 mole O © Total moles of product = 3+4 +42:6 +178.976 228.576

mole fraction of col= 3 28.57670-01312 - mole fraction of H₂O = 4 228.576 = 0.0175 nO 42.6 228. 576 mole fractions of Oz = 42 6 2 / 2 = 0.1864 mode fraction of Naz 178.976 = 0.183 = 0.1864 → - 178.976 :=0.783 228.576 - - - ----- Total mass of producte = 3844 + 4x18 t 42.6X32 +178.976 X 28 a (Comess) .. Ho mals mas) (Ne mall) = 6578.528 gram marte fraction of CO2 = mdf desx 44x3 6578.528

that free of 6 mooth*** mass fraction of col=0.0201 1884 mass Mass fraction of Hoa Hoa -50-0109 6578.528 mass mass fraction of O2 = 42.6432 - 6578-528 92-6x32 I 0.2072 mass fraction of N₂ = 178.976828 50.7618 6578.528