Experiment 4Recycling unfavorable economicspolluted environmentseparating recyclablesvirgin vs. recycledEnergy for AlAl more reactive than FeKeep volume of H2O smallAlum from Aluminum CanBring piece of AluminumMole 6.02x10232Al(s) + 2KOH(aq) + 4H2SO4(aq) + 22H2O(l)  2{KAl(SO4)2·12H2O}(s)+ 3H2(g) Add equationsLimiting Reagent1.4 molar Potassium hydroxide2Al(s) + 6H2O(l) + 2KOH(aq)  2K[Al(OH)4](aq) + 3H2(g)2K[Al(OH)4](aq) + H2SO4(aq)  2Al(OH)3(s) + K2SO4(aq) + 2H2O(l)Vacuum Filtration4. Calculate the number of moles present in each of the following. SHOW YOUR WORK!!2Al(s) + 2KOH(aq) + 4H2SO4(aq) + 22H2O(l)  2{KAl(SO4)2·12H2O}(s)+ 3H2(g) (a) 1.13 g of Al.1.13g/(26.9815g/mole) = 0.041880547 = 0.0419moles(b) 50. mL of 1.4 M KOH solution.50. mL * (1L/1000mL) = .050L0.050L * 1.4moles/L = 0.070moles(c) 20. mL of 9.0 M H2SO4 solution.20. mL * (1L/1000mL) = 0.020L0.020 L * 9.0 moles/L = 0.18 moles(d) 30. g (approximately 30. mL) of liquid water.30. g / 18.0152g/mole = 1.6652 = 1.7 moles5. (a) Using equation 9 in the CONCEPTS section and the amounts in question 4 (above), which reactantwill be the limiting reactant? Show your work!!!!! Use additional paper if necessary. .070 - 0.0419 = 0.0281 moles KOH left over.0419 * 2 = .0838 moles of sulfuric acid consumed.18 - .0838 = 0.0962 = 0.10 moles sulfuric acid remains22 moles water for 2 moles Al11 * 0.0419 = 0.461 moles water(b) How many moles of each of the non-limiting reactants will react?0.0419 moles KOH (c) How many moles of alum and H2 can theoretically be formed? (Assume a 100% yield.)(d) How many grams of alum can theoretically be formed?2C + O2  2CO2CO + O2  2CO22C + O2 + 2CO + O2  2CO + 2CO22C + 2O2  2CO2C + O2  CO22Al(s) + 2KOH(aq) + 4H2SO4(aq) + 22H2O(l)  2{KAl(SO4)2·12H2O}(s)+ 3H2(g) Actual YieldTheoretical Yield% YieldAt. Wt. of Al = 27.0g/moleMW Alum = 474g/moleIf you start with 1.000g of Al:1.000g/26.981538g/mole= 0.0370623 moles= 0.03706 moles of AlIn balanced equation, you have a 1:1 ratio of aluminum to alumTherefore the theoretical yield of alum would be 0.03706 moles of Alum for 1.000g aluminumMolecular Weight of Alum = 474.39g/moleTheoretical Yield of Alum = 474.39g/mole x 0.03706 moles =17.5820g = 17.58g Alum (your theoretical yield will be different based on how much aluminum you start with)Significant