(1) If 0.43g of aluminum foil was dissolved in sufficient 3M of KOH solution, an
ID: 780101 • Letter: #
Question
(1) If 0.43g of aluminum foil was dissolved in sufficient 3M of KOH solution, and the solution was acidified with 4M H2SO4, what is the theoretical yield of KAl(SO4)2 x 12H2O?
(2) If only 5.87g of alum were obtained as product from the synthesis described in (1), what was the percent yield?
(3) A student determined the Kf (f as lower subscript) of t-butyl alcohol using tap water instead of distilled or deionized water. Describe the problems that might have been encountered. How would these problems affect the magnitude of Kf?
Explanation / Answer
(0.43 g Al) / (26.9815 g Al/mol) x (1/1) x (474.3902 g KAl(SO4)2*12 H2O/mol) = 7.6 g KAl(SO4)2 * 12 H2O
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