Question

Problem Page Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 4.3 g of hexane is mixed with 26.8 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

CH₃ (CH₂) y CHz or C6H14 + 190₂ 6 Co₂ (g) + 7H₂O(g). Moles of hexane - Mass of hexane Molar mass of hexane 4. 3g z 0.05 mel. 86gmelt = Moles of O₂ = 26.88 32g|mel = 0.8375 mel. 2 first of all we have to find which reactant is. limiting : Ima of Cohly reacts with 19 moles of 02 10.05 meles of hexane reacts with 19 mele O2 x 0.05 mol 2 xl mel hexane Colly. = 0.475 mol O2 But there is 0.8375 ml of O2 ; O is present in excess ig mel of O2 reacts with Imole of CoHly 2 0.8375 mel ol o reacts with - Imel CG Hly x 0.8375 ml 19/mol O2 = 0.09 mal of Cohly. But there is only 0.05 mel of Colly, thertore hexane (Co Hly) acts as limiting reagent Amount of hexane will decide the amount of product formation.

I mol of hexane produces 0.05 mel of hexane will 7 mel H₂O x Imel C6Hly I me of water. produce :. 0.05 mel C6H14 = 0.35 mol H₂O. Mass of water - Moles & Molar mass = 0.35 me x 18 g mel Mass of water = 6.3 9.1