Asked by Melissa

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?

Answers

Answered by DrBob222
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. .
Answered by Anonymous
16.2%
Answered by Ui
Fcjjhvvb
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