A mixture contains only NaCl and Al2SO43 A 143g sample of th

A mixture contains only NaCl and Al₂(SO₄)₃. A 1.43-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 0.139 g. What is the mass percent of Al₂(SO₄)₃ in the sample?

Solution

Calculating moles of Al(OH)3 = 0.139 /78.0025=0.00178 moles

Moles of Al2(SO4)3 will be half that of Al(OH)3 i.e. = 0.00178/2 = 0.00089

Mass of Al2(SO4)3 for 0.00089 moles = 0.00089 * 342 = 0.30428 gms

% Al2(SO4)3 = 0.30428*100/1.43 = 21.29 %