Question

A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water, and an ex- cess 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?

Answer 1

Answer:

The mass percent of aluminum sulfate in the sample is 16.18%.

Explanation:

Mass of the sample = 1.45 g

$Al_2SO_3+6NaOH\rightarrow 2Al(OH)_3+3Na_2SO_4$

Mass of the precipitate = 0.107 g

Moles of aluminum hydroxide = $\frac{0.107 g}{78 g/mol}=0.001372 mol$

According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .

Then 0.001372 moles of aluminum hydroxide will be obtained from:

$\frac{1}{2}\times 0.001372 mol=0.000686 mol$

Mass of 0.000686 moles of aluminum sulfate :

= 0.000686 mol × 342 g/mol = 0.2346 g

The mass percent of aluminum sulfate in the sample:

$=\frac{ 0.2346 g}{1.45g}\times 100=16.18\%$