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EXAM #2 2. 20.0 g of AlzS3 and 20.0 g of H20 are reacted according to the following reaction: Al S; (s) + 6H20 (1) ► 2 Al(OH)3 (s) + 3 H2S (g) A. Determine the percent yield of the reaction if 17.5g Al(OH)3 are actually produced in this reaction. (8 pts) B. What mass of the excess reactant remains unused after the reaction in complete? (Bonus = 4 points)

Answer 1

Abszes) + 6H2Ole) 2Al(OH)3 (6) + 3 H₂S 19 on & mole of According to & prodand ALLOH)₂ Reaction & mole my 1 mole of I A6S3 Moo of Alz = 150.15 glmole mo of Alcon₂= 10 Ilmole mass taken of Also = 20 g mole of All - = 20 = 0.133 150115 According to reaction No. of moly prodund of recollz by 0.133 mol g Masa = 0: 2664 = Theoretical expected abtewned wit of ALCOH I = 20.77 = Theoretical yilld field. Ho ut obtained = 14.39 = Proctical field 1 yield - Practical yed - tivo Theoretical fild 84.25% =

moles of prodund risg of Alcol = w mod = 17.5 for formation of moles required = 0.22 4 moly 01224 moles of Alcoutz, of Al Sz = 0.112 mole But moles Thoufore perustnt of AldS3 2 0.133 remains moly = 0,133 – 0.112 = 0.02) so 0.021 moly of Alba will be unreacted unrearted = utamo = 0.021 x 150.15 wt I nost unreacted of Alas 3 = 3115 gm