Section: \\a Mass of aluminum used: .2785 12.5 Volume of 1.4 M KOH used: 1 the chille 1. mL Volume of 9.0 M H2SO4 used: 5.0 ortions c 'e the al ie solubi mL Mass of alum obtained: పిస39 % Yield (see questions below): procc 1." If y r using ed dirc Answer the following questions in the spaccs providcd. To carn complete credit show all your work completely and neatly. Give the chcmical formula and molar mass of the alum produccd in this investigation. 1. conta KAICSOP 12H20 sink. mm = 474 1g Write answer to prelab question number 4 here after you compare your reaction with 2. reaction (5) from the "CONCEPTS OF THE EXPERIMENT" section. 2al)+22H20@.aKOHlad) t 40t SO109→ 34,6)+2K aIsola 24.0) GHociao far Cale Calculate the number of moles of the chemical spccified in each of the following: 3. (a) H2SO4 in 5.0 mL of 9.0 M H2SO4 solution. Oi045Moles om (b) KOH in 12.5 mL of 1.4 M KOH solution. O.0175 moles (c) Water in 10 mL (about 10 g) of water. 0.535 moles (d) Aluminum present in the mass of aluminum you used in this experiment. 7859 26.98 O,0103 moles Glum used in expeniment 86 grva wnite Slr! ma (lear' Chan 3 lie AICI3 milky Whip e i Na2SO4 83 i number 3 above, which reactant will be the limiting rcactant? See atachod Cale from question 2 above and the numbers of moles calculated in komuAIMIle. ai is limiting (b) How many moles of alum and H2 can theorctically (assume 100 % yield) be formcd? O, 0103 mores Qlum-2=0.02016 O.0154s moles Ha Preduce 3 moles Ho s alum molos alum moles alum. Droduco moles alum VOur po any grams of alum can thcoretically be formed? Based on this number and your actual yield, what is the percentage yield of alum produccd in your experiment? whatwegoi inexp Act y % yield X 100 Theor y th max amount I couldue producoo always basedon limtina react, 3.93% 3.2739 0.01020 List at least two causes for a yicld greatcr than 100 %. 5. Of the ineasued product of the rxn containes impurities that cause its mass tobe itwould ifit waspune. greaken than 2) DUsed 6. more readlant than youthought Hum How many grams of alum can be obtained from 20.0 g of aluminum when the reaction proceeds with 100 % yield? (assume the other rcactants are present in excess) How C many grams of alum would be obtained if the reaction were to proceed with the yield achieved in your experiment? hoo vield 132 1g 279/m moles CMass -0.037mol atmas (sec the next page) 87