Question

-Lab_A for Synthesis The Synthesis of Alum DATA TABLE Mass of Al Used 116 40,5607 Mass of empty watch glass Mass of watch glass and Alum sample Mass of Alum Obtained Show calculations and/or discuss the following in your lab report Determine the theoretical yield of the alum. Use the aluminum foil as the limiting reagent and presume that the foil was pure aluminum. 1. 2. Calculate the percent yield of your alum crystals. 3. Discuss the factors that affected the percent yield. 4. Describe 2 uses of alum,

Answer 1

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

                                  = 1.1649 g / 26.982 g mol^-1 = 0.04317 moles

# 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)  

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 = 0.04317 moles

# MW of alum KAl(SO₄)₂^.12H₂O = 474.3904 g/ mol

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

                                    = 0.04317 moles x 474.3904 g mol^-1

                                    = 20.479 g

Hence, theoretical yield = 20.479 g

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

                        = (17.7066 g / 20.479 g) x 100

                        = 86.46 %    

#3. The factors affecting % yield are-

I. Incomplete precipitation and loss of Alum during filtration decreases % yield.

II. The reaction takes place in acidic medium. So, presence of inadequate amount of H₂SO₄ (i.e. not taken in excess) also decreases % yield.

#4. Uses of Alum:

I. Alum is used as flocculants to increase the apparent mass of solvated particles and remove them from the liquid/water for purification purpose.

II. Use in preparation of cosmetics including after-shave lotions, deodorants, skin toner, etc.