Question

After dissolving 1.430g of a mixture of Na₂SO₄ and Al₂(SO₄)₃ in water, you add an excess of BaCl₂ solution. You filter off the preciptitate which forms, dry it, and find that it weighs 2.54g. What is the percent by mass of Al₂(SO₄)₃ in the original mixture?

Answer 1

Balanced chemical reaction take place is

Al₂(SO₄)_3(aq) + 3 BaCl_2(aq) 3 BaSO_4(s) + 2AlCl_3(aq)

in reaction precipitate of BaSO₄ formed

molar mass of BaSO₄ = 233.38 gm/mole then 2.54 mole of BaSO₄ = 2.54 / 233.38 = 0.01088 mole

According to reaction 1 mole of Al₂(SO₄)₃ form 3 mole of BaSO₄ then to form 0.01088 mole of BaSO₄ required Al₂(SO₄)₃ = 0.01088 / 3 = 0.0036278 mole

Al₂(SO₄)₃ present in mixture = 0.0036278 mole

molar mass of Al₂(SO₄)₃ = 342.15 gm/mole then 0.0036278 mole of Al₂(SO₄)₃ = 0.0036278 X 342.15 = 1.241267 gm

Al₂(SO₄)₃ present in mixture = 1.241267 gm

1.430 gm mixture = 100 % then

1.241267 gm Al₂(SO₄)₃ = 1.241267 X100 / 1.430 = 86.80 %

percent by mass of Al₂(SO₄)₃ in original mixture = 86.80 %