Alum Synthesis Pre-Lab CHM205 Laboratory Spring 2018 Section: Grade Nam A studen

Question

Alum Synthesis Pre-Lab CHM205 Laboratory Spring 2018 Section: Grade Nam A student prepared a sample of alum us He started with 0.7 ing the same method that you will perform in lab this wee 88 g of aluminum fol, and after drying for a week, obtained 11.187 gof alum. 1. Calculate the number of moles of aluminum the student started with 2. Write the balanced chemical equation for aluminum foil reacting with potassium hydroxide an sulfuric acid reacting to produce alum (including the 12 waters of hydration) Using the stoichiometry from the overall equation for the synthesis of alum.and ho d be produced that aluminum foil is the limiting reagent, determine how many moles in this reaction. nd the assumption m should be produced 3. of alu Calculate the molar mass of alum (including the 12 waters of hydration!). 5. Calculate the theoretical yield of alum in grams for the reaction that the student performed. 6. Calculate the percent yield of alum that the student obtained. Lab #7: Alum Synthe Page l69

Explanation / Answer

Ans. #1. Moles of Al = Mass of Al metal / atomic mass of Al

                                  = 0.788 g / 26.982 g mol-1 = 0.0292 moles

#2. Balanced reaction for Alum synthesis:

2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 22 H2O(l) ----> 2 KAl(SO4)2.12H2O + 3 H2(g)  

#3. From stoichiometry of the reaction, 1 mol Al forms 1 mol alum. The formation of product follows the stoichiometry of limiting reactant (here, Al).

Thus, (theoretical) moles of alum produced = 0.0292 moles

#4. MW of alum KAl(SO4)2.12H2O = 474.3904 g/ mol

#5. Theoretical mass of alum produced = Moles of alum x MW alum

                                    = 0.0292 moles x 474.3904 g mol-1

                                    = 13.854 g

#6. % yield = (Actual yield / theoretical yield) x 100

Use the actual yield (experimental yield) obtained in the experiment to calculate % yield.

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