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Alum [KAl(SO4)2·xH2O] is used in food preparation, dye fixation, and water purification. To prepare alum, aluminum...

Alum [KAl(SO4)2·xH2O] is used in food preparation, dye fixation, and water purification. To prepare alum, aluminum is reacted with potassium hydroxide and the product with sulfuric acid. Upon cooling, alum crystallizes from the solution.

(a) A 0.4711−g sample of alum is heated to drive off the waters of hydration, and the resulting KAl(SO4)2 weighs 0.2564 g. Determine the value of x and the complete formula of alum.

Value of x:

Complete Formula:

(b) When 0.7339 g of aluminum is used, 8.415 g of alum forms. What is the percent yield?

% yield=

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Answer #1

a) Mass water = 0.4711 - 0.2564 = 0.2147g
Moles water = 0.2147 / 18.02 = 0.01191
Molar mass KAl(SO4)2 = 258.205 g/mol
Moles KAl(SO4)2 = 0.2564 g / 258.205 = 0.0009930
We get
KAl(SO4)2 ( 0.0009930) H2O ( 0.01191)
We divide by the smallest number : 0.01191 / 0.0009930 = 11.9939
x = 11.9939 and the formula is KAl(SO4)2 . 11.9939 H2O

b) 0.7339 x 100 / 8.415 = 8.72133 %

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