Al2(SO4)3(aq) + BaCl2(aq) BaSO4(s) + AlCl3(aq)

Consider the unbalanced equation above. A 0.680 g sample of impure Al2(SO4)3 reacts with excess BaCl2. If the sample produces 0.530 g of BaSO4, what is the mass percent of Al2(SO4)3 in the sample?

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

First, let's determine the number of moles of BaSO4 produced. We can use the molar mass of BaSO4 to convert the mass of BaSO4 to moles.

Molar mass of BaSO4:
Ba = 137.33 g/mol
S = 32.06 g/mol (Note: There are 4 S atoms in BaSO4)
O = 16.00 g/mol (Note: There are 4 O atoms in BaSO4)

Molar mass of BaSO4 = (137.33 g/mol) + (4 * (32.06g/mol)) + (4 * (16.00 g/mol)) = 233.40 g/mol

Now, let's calculate the number of moles of BaSO4:
moles of BaSO4 = 0.530 g / 233.40 g/mol = 0.00227 mol

Next, we need to use the balanced equation to determine the molar ratio between Al2(SO4)3 and BaSO4. According to the equation, 1 mole of BaSO4 is produced for every 1 mole of Al2(SO4)3. Therefore, the number of moles of Al2(SO4)3 is the same as the number of moles of BaSO4.

So, moles of Al2(SO4)3 = 0.00227 mol

Now, we can find the molar mass of Al2(SO4)3:
Al = 26.98 g/mol (Note: There are 2 Al atoms in Al2(SO4)3)
S = 32.06 g/mol (Note: There are 3 S atoms in Al2(SO4)3)
O = 16.00 g/mol (Note: There are 12 O atoms in Al2(SO4)3)

Molar mass of Al2(SO4)3 = (2 * (26.98 g/mol)) + (3 * (32.06g/mol)) + (12 * (16.00 g/mol)) = 342.15 g/mol

Finally, let's calculate the mass percent of Al2(SO4)3 in the sample:
mass percent of Al2(SO4)3 = (moles of Al2(SO4)3 * molar mass of Al2(SO4)3) / mass of sample) * 100%

mass of sample = 0.680 g

mass percent of Al2(SO4)3 = (0.00227 mol * 342.15 g/mol) / 0.680 g) * 100% = 11.44%