Question

38) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.0 g of HCl according to the following equation. Al2O3(s) + 6HCl(aq) + 2AlCl3(aq) + 3H2(aq) A) 16.28 B) 20,0 g C) 12.28 D) 10.08 E) 6.10 g

Answer 1

AlO3 (6) + 6H (ag) 2 All + 2H20 Lag) Weight of AlO3 = 10.og. no. of mole of Al₂O₃ = Weight = 10.09 = 0.098 map Melar mas 10 2ginal = 0.1 mol weight of the = 10.ug. no. of male of the = 10.og = 0.274 mol 36.5 g mel Ace to sx 1 mol Aloz react with 6 mol Hle, then bilmed Al Oz react with 0.6 mol the but we have only 0.274 mol re. Hence Hill is limiting readard 0.274 mol he react with Ix 0.274 = 0.04567 mol Alo Xo 6 mol He give 2 mol Alces alc. to an then 0.274 mol HCl give ? x 0,274 = 0.09133 mol All mais of Alche = nxmolar mais = 0.09133 mulx 133.34 g/mol = 12.2g Hence, 12.2g of Alcez produced Any cy 12.29.