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?

See your earlier post below.

93.15%

To find the mass percent of Al2(SO4)3 in the sample, we need to determine the mass of Al2(SO4)3 in the precipitate and the total mass of the sample.

1. First, let's calculate the moles of Al(OH)3 precipitate formed. We know the mass of the precipitate is 0.126 g, and we can find its molar mass.

The molar mass of Al(OH)3 is:
Al: 1 atom x 26.98 g/mol = 26.98 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
H: 9 atoms x 1.01 g/mol = 9.09 g/mol

Adding these up, we get:
26.98 g/mol + 48.00 g/mol + 9.09 g/mol = 83.07 g/mol

Now we can calculate the moles of Al(OH)3 precipitate:
0.126 g / 83.07 g/mol ≈ 0.00152 mol

2. The equation for the reaction between Al2(SO4)3 and NaOH is as follows:
Al2(SO4)3 + 6NaOH → 2Al(OH)3 + 3Na2SO4

From the balanced equation, we can see that the mole ratio between Al2(SO4)3 and Al(OH)3 is 1:2. Therefore, the moles of Al2(SO4)3 in the precipitate is half of the moles of Al(OH)3.

Moles of Al2(SO4)3 in precipitate = 0.00152 mol / 2 ≈ 0.00076 mol

3. Now, to find the mass of Al2(SO4)3 in the sample, we can calculate its molar mass:
Al: 2 atoms x 26.98 g/mol = 53.96 g/mol
S: 1 atom x 32.07 g/mol = 32.07 g/mol
O: 12 atoms x 16.00 g/mol = 192.00 g/mol

Adding these up, we get:
53.96 g/mol + 32.07 g/mol + 192.00 g/mol = 277.03 g/mol

Mass of Al2(SO4)3 in sample = 0.00076 mol x 277.03 g/mol ≈ 0.209 g

4. Finally, we can calculate the mass percent of Al2(SO4)3 in the sample:
Mass percent = (mass of Al2(SO4)3 / total mass of sample) x 100

Total mass of sample = 1.76 g

Mass percent of Al2(SO4)3 = (0.209 g / 1.76 g) x 100 ≈ 11.9%

Therefore, the mass percent of Al2(SO4)3 in the sample is approximately 11.9%.