250 mL of 0.18M potassium sulfate solution is added to a 100 mL solution of 0.25 M barium nitrate. Calculate the mass of the precipitate formed.

Question

Doing a year long review and forgot the equations for these, a walkthrough and the equations would be greatly appreciated.

Answer 1

1) Determine the reaction products:

K₂SO₄ (aq) + Ba(NO₃)₂ (aq) ---> KNO₃ (aq) + BaSO₄ (s)

2) Balance the equation:

K₂SO₄ (aq) + Ba(NO₃)₂ (aq) ---> 2KNO₃ (aq) + BaSO₄ (s)

3) Determine the moles for each reactant:

K₂SO₄ (aq)

M = mol/L

0.18 = mol/0.25

0.18 (0.25) = mol

0.045 = mol

Ba(NO₃)₂ (aq):

M = mol/L

0.25 = mol/0.10

0.25 (0.10) = mol

0.025 = mol

4) Determine the limiting reactant:

0.045 mol K₂SO₄ x (1 mol BaSO₄ / 1 mol K₂SO₄) = 0.045 mol BaSO₄

0.025 mol Ba(NO₃)₂ x (1 mol BaSO₄ / 1 mol Ba(NO₃)₂) = 0.025 mol BaSO₄

Limiting reactant = Ba(NO₃)₂

5) Use limiting reactant to determine mass of precipitate:

0.025 mol BaSO₄ x (233.38g / mol) = 5.83g BaSO₄

6) Answer:

Mass of precipitate = 5.83g BaSO₄