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Experiment 4 Recycling unfavorable economics polluted environment separating recyclables virgin vs recycled Energy for Al Al more reactive than Fe Keep volume of H2O small Alum from Aluminum Can Bring piece of Aluminum Mole 6 02x1023 2Al s 2KOH aq 4H2SO4 aq 22H2O l 2 KAl SO4 2 12H2O s 3H2 g Add equations Limiting Reagent 1 4 molar Potassium hydroxide 2Al 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 Filtration 4 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 050L 0 050L 1 4moles L 0 070moles c 20 mL of 9 0 M H2SO4 solution 20 mL 1L 1000mL 0 020L 0 020 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 moles 5 a Using equation 9 in the CONCEPTS section and the amounts in question 4 above which reactant will 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 remains 22 moles water for 2 moles Al 11 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 2CO 2CO O2 2CO2 2C O2 2CO O2 2CO 2CO2 2C 2O2 2CO2 C O2 CO2 2Al s 2KOH aq 4H2SO4 aq 22H2O l 2 KAl SO4 2 12H2O s 3H2 g Actual Yield Theoretical Yield Yield At Wt of Al 27 0g mole MW Alum 474g mole If you start with 1 000g of Al 1 000g 26 981538g mole 0 0370623 moles 0 03706 moles of Al In balanced equation you have a 1 1 ratio of aluminum to alum Therefore the theoretical yield of alum would be 0 03706 moles of Alum for 1 000g aluminum Molecular Weight of Alum 474 39g mole Theoretical Yield of Alum 474 39g



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