Question

A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of Al(OH)3. The precipitate is filtered, dried, and weighed. The mass of the precipitate is 0.107 g. What is the mass percent of Al2(SO4)3 in the sample?
Al2(SO4)3 + 6NaOH ==> 2Al(OH)3 +3Na2SO4

0.107 g Al(OH)3 x [1 mole Al2(SO4)3/2 moles Al(OH)3] = ??g Al2(SO4)3
Convert to percent b7 dividing by 1.45 g and multiplying by 100. .
16.2%
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