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 mass of Al2(SO4)3 present in the 0.680 g sample.

First, we need to determine the molar mass of Al2(SO4)3:
Al: 2 * 26.98 g/mol = 53.96 g/mol
S: 3 * 32.07 g/mol = 96.21 g/mol
O: 12 * 16.00 g/mol = 192.00 g/mol

Molar mass of Al2(SO4)3 = 53.96 + 96.21 + 192.00 = 342.17 g/mol

Next, we can use stoichiometry to relate the mass of BaSO4 produced to the mass of Al2(SO4)3 reacted.

From the balanced chemical equation, we can see that the ratio of moles of BaSO4 to moles of Al2(SO4)3 is 1:1.

Mass of BaSO4 = 0.530 g

Using the molar mass of BaSO4:
Molar mass of BaSO4 = 137.33 g/mol

Now, we can calculate the moles of BaSO4 produced:
Moles of BaSO4 = Mass of BaSO4 / Molar mass of BaSO4
= 0.530 g / 137.33 g/mol

Since the ratio of BaSO4 to Al2(SO4)3 is 1:1, the moles of BaSO4 produced would be equal to the moles of Al2(SO4)3 reacted.

Next, we can calculate the moles of Al2(SO4)3:
Moles of Al2(SO4)3 = Moles of BaSO4

Then, we can calculate the mass of Al2(SO4)3:
Mass of Al2(SO4)3 = Moles of Al2(SO4)3 * Molar mass of Al2(SO4)3
= Moles of BaSO4 * Molar mass of Al2(SO4)3
= (0.530 g / 137.33 g/mol) * 342.17 g/mol

Finally, we can determine the mass percent of Al2(SO4)3 in the sample:
Mass percent of Al2(SO4)3 = (Mass of Al2(SO4)3 / Mass of the sample) * 100%
= (0.530 g / 0.680 g) * 100%

To find the mass percent of Al2(SO4)3 in the sample, we need to calculate the mass of Al2(SO4)3 in the 0.680 g sample.

1. Calculate the molar mass of BaSO4:
Ba = 137.33 g/mol
S = 32.07 g/mol
O = 16.00 g/mol

Molar mass of BaSO4 = (137.33 g/mol) + (32.07 g/mol x 4) + (16.00 g/mol x 4) = 233.38 g/mol

2. Calculate the moles of BaSO4 produced:
Moles = Mass / Molar mass
Moles of BaSO4 = 0.530 g / 233.38 g/mol = 0.00227 mol

3. According to the balanced equation, the stoichiometric ratio between Al2(SO4)3 and BaSO4 is 1:1. That means for every 1 mol of BaSO4 produced, 1 mol of Al2(SO4)3 reacts.

4. Calculate the moles of Al2(SO4)3 in the sample:
Moles of Al2(SO4)3 = 0.00227 mol

5. Calculate the molar mass of Al2(SO4)3:
Al = 26.98 g/mol
S = 32.07 g/mol
O = 16.00 g/mol

Molar mass of Al2(SO4)3 = (26.98 g/mol x 2) + (32.07 g/mol x 3 x 4) + (16.00 g/mol x 3 x 4) = 342.18 g/mol

6. Calculate the mass of Al2(SO4)3 in the sample:
Mass = Moles x Molar mass
Mass of Al2(SO4)3 = 0.00227 mol x 342.18 g/mol = 0.777 g

7. Calculate the mass percent of Al2(SO4)3 in the sample:
Mass percent = (Mass of Al2(SO4)3 / Mass of sample) x 100%
Mass percent of Al2(SO4)3 = (0.777 g / 0.680 g) x 100% = 114.26%

The mass percent of Al2(SO4)3 in the sample is approximately 114.26%.