Homework Help: chemisty

Posted by hope on Saturday, February 5, 2011 at 10:58pm.

A mixture contains only NaCl and Al2(SO4)3. A 1.76-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 0.126 g. What is the mass percent of Al2(SO4)3 in the sample?

chemisty - DrBob222, Saturday, February 5, 2011 at 11:33pm
%Al2(SO4)3 = (mass Al2(SO4)3/mass sample)*100 = ??

How to find mass Al2(SO4)3.
You have mass Al(OH)3 = 0.126 g. Convert to mass Al2(SO4)3.
How many moles is that?
moles Al(OH)3 = grams/molar mass = ??

There are 2 moles Al in 1 mole Al2(SO4)3; therefore, moles Al2(SO4)3 = moles Al(OH)3 x (1 mole Al2SO4)3/2 moles Al(OH)3) and ?? moles Al(OH)3 x (2/1) = moles Al2(SO4)3.
Convert moles Al2(SO4)3 to grams. g - moles x molar mass and substitute into the top equation.

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