Question

Name Instructor balanted: 2A1s) 2.4DMg) 4H5t 229400-2KAI()12420 43W2ug Date EXPERIMENT REPORT SHEET Recycling Aluminum Experimental Data: Synthesis of KAKSO-12 H glass ler Poper 38. 2010 Mass of Al Volume of KOH Wotch 15 mL 5.5M mL Volume of H,SO, Mass of product 43.297 .201 Calculations Moles of Al 0141mol Moles of KOH Moles of H SO Moles of product (KAKSO 12HO) Limiting reagent Theoretical yield (KAISO-12 H,O) Percent yield (KA(SO-12 H,O) n d d w ct Comment on possible impurites in your product Show all work for determining the percent yield whte frecqtore kmed han ock Oddec COnhio Sulfote.Copright O 2015 Pearson Education, Inc. 96

Answer 1

Chemistry to produce 8.0242 moles of Kallsoul. 12 10 T 0.03615 males of H. 2 pels) +2 kohkaq) + 4 4,50 Caq 2+22 Ho lo the It can be seen from eph @ 0 KOH $ H₂SO4 is available in exuss reactant is Allo. And limitting 2 KAL (S06)2 - 1240 + 3H₂(g) moles of product Ck al? Sow : 2420) = Met = 0.01 given: Mass of AL = 0.651 gm Volume of KaM = 25 mL = 0·025h valume of Hart = 15.5 mh a 0.0155L mass of product = 5:0ggm moles of al = 0.0241 male - ③ Moles of kon = 0.05 mol - Molas of sou = 0.093 mele - 52000000000 When 100% of limiting reactant Thoritical Yield = Maximum amount of product Cam) which can be gut from the reaction. Actual Yield Amount of productlynly thich is actually generating - Actual Vield Tharitical field ,x100 Percentage field En 1 can also be written as Actual yield = 5.0ggm Thoritical Yield = 0.0241x474.39 = 11.439m max amount of most of produse considering Kreis94) 12,0 100% Convilom +!!MO ) 184,50CA K ACS00)... 12H0+49) Alis) + KDH of Alko 1 mole At is) can be react with Imole koha 2 mow H₂S & Imoles to 96.52% Am % Yield 5.08 100 = T.43 1 mol al & 2 mole, Kon = amore H₂ SO, Z Il mole H₂O There fore 0.024 1 mole Ad (s) required 0.024 1 molay Konto.0482 moles of H₂SO4 So. 265 h moles of mater