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? The first person who answered this was wrong. The answer is supposed to be 33% but can someone show me how?

Answer 1

Molar masses:
Na2SO4 - 142.04 g/mol
Al2(SO4)3 - 342.15 g/mol
BaSO4 - 233.39 g/mol

Assume that the amount of Na2SO4 in mixture is x

Amount of Al2(SO4)3 in the mixture. y
x+y = 1.430 g   ----1

Reaction;

SO4-2 + Ba++2 = BaSO4

Need total number of SO4-2

Number of mole = amount in g/ molar mass

Moles of SO4-2 from Na2SO4:
x / (142.04 g/mol)

Moles of SO4-2 from Al2(SO4)3:
3y / (342.15 g/mol)

Moles of SO4-2 from BaSO4:
2.540 g / (233.39 g/mol) = 0.010883 mol

x / (142.04 g/mol) + 3y / (342.15 g/mol) = 0.010883 mol ----2

to solve equation number 1 anad 2;

(1.430 - y)/142.04 + 3y/342.15 = 0.010883
0.010068 - 0.0070403y + 0.0087681x = 0.010883
0.0017278x = 8.15e-4

y=0.472

% = amount of Al2(SO4)3/ amount of mixture]*100

= 0.472/ 1.430 ]*100

= 33.0% Al2(SO4)3 by massMolar masses:

Na2SO4 - 142.04 g/mol
Al2(SO4)3 - 342.15 g/mol
BaSO4 - 233.39 g/mol

Assume that the amount of Na2SO4 in mixture is x

Amount of Al2(SO4)3 in the mixture. y
x+y = 1.430 g   ----1

Reaction;

SO4-2 + Ba++2 = BaSO4

Need total number of SO4-2

Number of mole = amount in g/ molar mass

Moles of SO4-2 from Na2SO4:
x / (142.04 g/mol)

Moles of SO4-2 from Al2(SO4)3:
3y / (342.15 g/mol)

Moles of SO4-2 from BaSO4:
2.540 g / (233.39 g/mol) = 0.010883 mol

x / (142.04 g/mol) + 3y / (342.15 g/mol) = 0.010883 mol ----2

to solve equation number 1 anad 2;

(1.430 - y)/142.04 + 3y/342.15 = 0.010883
0.010068 - 0.0070403y + 0.0087681x = 0.010883
0.0017278x = 8.15e-4

y=0.472

% = amount of Al2(SO4)3/ amount of mixture]*100

= 0.472/ 1.430 ]*100

= 33.0% Al2(SO4)3 by mass