Question

During a "Synthesis of Alum" experiment, with 24.9mL of 1.4M KOH. After cooling the mixt Given the following balanced equation: OF Alum" experiment, a student uses 0.500g of aluminum foil and disson NOR. After cooling the mixture, the students adds 11.9mL of 9.OM H2SO4. 2 Al(s) + 2 KOH(aq) + 4 H,SO(aq) + 22 H200) — 3 H2(g) + 2 KAl(SO4)2-12 H20 a. Calculate the theoretical yield of alum. Show the process of finding the limiting reactant. [Molar Mass (Alum) = 474.3g/mol Alum) (-15 points) b. If the student isolates 6.502g of alum as crude product, calculate the percent yield. [5 points)

Answer 1

The linen balanced chemical reaction is ; d4116) + d Kol (aq) + 4H, soy laq) +22 Ho res — 34449) t« KAI (S04), 1240. from the linen chemical reaction, we can say that two moles of Al reacts with a moles of KOH and 4 moles of H₂SO4 . And gives a moles of Alum. Thes, chemical reaction total gines relationship in terms of moles. This implies we need to calculate the noles 4 Reactent from the prouided information * 0.5eogą Al foil # moles of Al - Gramof Al deseting Atonic Mars #moles of Al 2 0.500g “: 27 gmol #moles Al = 0.018 mele t o # 2409 ml of 1.4M KOM. uber of moles y 1000 Molarity a member of moles 0 wolume du mL → number of moles : Molarity & noluine in ml Holarity 21.4M 1000 h volume a 94.9 ml :: number of moles. > 4.4x24.9 muneberit moles 04 KOH = 0.0347» @ 1000

© 11.9 mi 1 9.0 M. Hyson Molavity' = 9.0M volume z 11.9ml number of moles (9.0M) x 11.9 mL looo - 0.1070 moles. 1 # 145o = 00/07 moles to I moles of Al require & moles of KOH & 2 moles of thSou 0.018 moles of Al require 0.018 moles of KOH & 0.036 moles of happy. her moles of kon we moles of Al we have = 0.034 O moles of Hy sou have af 0.019. We hau 20.107. 3) Al get consumed 100% in the leaction » Al is the limiting reagent og al mole of Al produced 2 moles 4 Alum. $ 0.018 moles of At produced 0.018 moles of Alum. #moles g of Alum prockeced Molai mers of Alum grave of Alum ploewed z #motesx Molai mars 4 Alem = 10.018) X ( 494. 3) -8.537 g (Theoretical field 4 Alum z 8.53797 Answer %oge yield > Experiment yield X100% Theoretical yield = 6.50 29 x 100 % = 176.16 % + Awesen