Question

Determine the mass of precipitate, in grams, that forms when 56.4 ml, of 0.0354 MBa(CIO4)2 reacts with 69.7 mL of 0.0591 MK2SO4. 2KCIO4(ad)+ BaSO4(s) Ba(CIO4)2(ac+ K2SO4(a) g precipitate

Answer 1

The given balanced reaction is

Ba(C104)2(aq) + K2SO4(aq) + 2KCIO4(aq) + BaSO4(s)

Total volume of solution taken = V = 56.4 mL = 0.0564 L

BaC1O4)2(aq)

Concentration of solution, C = 0.0354 M = 0.0354 mol/L

BaC1O4)2(aq)

Hence, the number of moles of can be calculated as

BaC1O4)2(aq)

0.0354 mol CXV=- x 0.0564 L 1 L 0.001996 mol

Total volume of K₂SO₄ taken = V = 69.7 mL = 0.0697 L

Concentration of K₂SO₄ = 0.0591 M = 0.0591 mol/L

Hence, the number of moles of K₂SO₄ taken can be calculated as

CXV=- 0.0591 mol x 0.0697 L 1 L 0.004119 mol

From the balanced equation, 1 mol of
BaC1O4)2(aq)
reacts with 2 moles of K₂SO₄ , Hence, the number of moles of K₂SO₄ that will react with 0.001996 moles of
BaC1O4)2(aq)
can be calculated as

2 mol K2SO4 - X 0.001996 mol Ba(ClO4)2 = 0.003993 mol K2SO4 1 mol Ba(C/O)2

Since we have more K₂SO₄ than the required amount that reacts, the limiting reactant is .

BaC1O4)2(aq)

Hence, the number of moles of precipitate that forms will depend on the number of moles of .

BaC1O4)2(aq)

From the balanced reaction, the number of moles of precipitate that forms can be calculated as

1 mol BaSO4 – X 0.001996 mol Ba(ClO4)2 = 0.001996 mol BaSO4 1 mol Ba(C/O)2

Molar mass of BaSO₄ precipitate = 233.38 g/mol

Hence, the mass of precipitate formed can be calculated as

mass = moles x molar mass = 0.001996 mol x 233.38 50.466g

Hence, the amount of precipitate that forms is 0.466 g.