Question

0.189 moles of ethane gas are mixed with 2.1 moles of oxygen in a sealed container and the mixture is then ignited (initiating a combustion reaction) to form carbon dioxide and water. 2 C2H6(g) + 702(g) 4CO2(g) + 6H2O(g) What is the volume of CO2 formed when the partial pressure of CO2 is 200 kPa and a temperature of 200 °C? (Assume that all gases behave ideally). (A) (B) (C) (D) (E) 7.4 x 10-3 m3 3.7* 10-3 m3 4.1 x 10-2 m3 7.4 m3 2.1 x 10-2 m3

Answer 1

Sol. ² Given, moles of GH(I) = 0.189 moles 0,19) = 2.1 moles. Balanced combustion reaction, a C2H6 (3) +7(9) —> 4COL (9) + 6 H10 (9) now, To find limiting reagent, (divide each readants moles by its respective Cebficient) <GH> = 0.199 = 0.0945 CO2) = 1 = 0.300 - <LH) is less than <O₂) - Cho is limiting reagent. now - As per stoichiometry of above reaction, 2 mol CatHo produce 4 mol co,. 0.189 mol Ca Ho will produce, → 4x0.189 mol CO2 1 = 0.378 mol CO₂. So, med Neon = 0.378 mol. CamScanner

; Peon = 200 kPa = 200 x.to3 Pa T = 200°C = 473 R. - q dead - Ideal gas constant, R = 8.3145 m3. Pa K. mol gas equation, PV = nRT where, P= Pressure VVolume na no. of moles R= Ideal gas Constant T = Temperature 5 V= nRT p. - 0.378 mol x 8.3145 m. Pa x 473k komol 200 x 103 Po = v= 7.4 x 10-3 m3 Sol Option CAD is correct. Scanned with CamScanner