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A mixture contains only NaCl and Al2(SO4)3. A 1.75-g sample of the mixture is dissolved in...

A mixture contains only NaCl and Al2(SO4)3. A 1.75-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 0.134 g. What is the mass percent of Al2(SO4)3 in the sample?

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Answer #1

It is given precipitate = Al(OH)3

Mass of precipitate, Al(OH)3 = 0.134 g

moles of precipitate formed =

Al2(SO4)3 + 6 NaOH --> 2 Al(OH)3 & 3 Na2SO4

From the question it is clear that for the mixture of Al2(SO4)3 & NaCl, the reaction shows that NaOH reacts with Al2(SO4)3 to form Al(OH)3

Mass of Al2(SO4)3 required =

Mass of Al2(SO4)3 required to produce 0.134 g Al(OH)3 = 0.294 g Al2(SO4)3

Mass percentage of Al2(SO4)3 is =

% mass of Al2(SO4)3 = 16.8

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