Question

Answer # 14 15. If 2.0 moles of octane vapor, C8H18, is reacted with 15 moles of oxygen to produce carbon dioxide and water, what are the partial pressures of all gases in the mixture at a total pressure of 1 atm? PH20 6Sk) atm mswer #15: PCSH18= 5.. Propane (C3H8) gas is used in gas grills and is combusted with oxygen to produce carbon dioxide and water vapor. If 5.00 L of propane is combusted, what volume of oxygen is required and what volumes of water vapor and carbon dioxide are produced? Pco2 =0 atm Poz= atm atm

Answer 1

Org ron Limiting Reagent IS 25/2 - 1.2 12.5 ncoz 15. Cg Hig = 2 mol O2 = 15 moles. Protell = 1 atm Cg H180 + 256 - 8202 + 9 H2OCO - moles 2 is Limiting 2 > I aisla - Go will act as limiting Reagent (hCGH18) used = nou a negtig = 1 s os =12 (hC8 H18) remained = 2-12 -9 0-8 moles no2 nco2 = 8x 15 = MCQ = 9.6mol 8 12.5 hoz = nH₂O= 9x15 nH₂O = 10.8mo) - Peghig = neghig Protel I) 1.2 xl PC & Hig= 0.0556 atna Pcos = nco, * Protel 9.6 x xl 7 Peq = 0.44 atm/ hag Hug theo, thr20 21 - HH₂O x Ptotul » 10 x1 7 PH₂O = 0.5 atm 10 naghig theo, theo O2 is consumel completely As so Poq = 0 atm 12.5 WHO - 12.5 12- 5 1.2 + 906 +10.8 hag His the cotheo PH2O = negligt 2)-6

is. Vezug = slit (3 48 (+ 5O2(g) VigHB = Voz a 3602 og UH2O (6) Voz = sxs voz = 25 cit An VCO2 = 345 coz = is lit Ane VizuB - VGOD =) VigHB = VH₂O = VHO = uxs WHO = 20 lit And

Answer 2

Calculate the rate of burning of carbon particle 2.56 cm radius in an atmosphere of pure oxygen at 1000 Kand 1 atm. Assuming a very large blanking layer of COz has formed a round the particle. At the carbon surface Pco2=1 atm and Poz=0. At very large radius Pcoz=0 and Poz=1 atm. Given the diffusivity of oxygen in carbon dioxide =1.032 cm2/s.