Question

A 20.0mL sample of a 2.00M potassium sulfate solution is mixed with 14.8mL of a 0.880M barium nitrate solution and this precipitation reaction occurs:

K2SO4(aq)+Ba(NO3)2(aq)?BaSO4(s)+2KNO3(aq)

What is the theoretical yield?

Answer 1

Number of mmol of potassium sulfate is , n₁ = Molarity x volume in mL

                                                                = 2.00 M x 20.0 mL

                                                                = 40.0 mmol

Number of mmol of barium nitrate is , n₂ = Molarity x volume in mL

                                                           = 0.880 M x 14.8 mL

                                                           = 13.02 mmol

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

Accordig to the balanced equation ,

1 mole of K₂SO₄ reacts with 1 mole of Ba(NO₃)₂

                               OR

1 mmol of K₂SO₄ reacts with 1 mmol of Ba(NO₃)₂

13.02 mmol of K₂SO₄ reacts with 13.02 mmol of Ba(NO₃)₂

So 40.0 - 13.02 = 26.98 mmol of K₂SO₄ left unreacted.so Ba(NO₃)₂ is the limiting reactant.

from the equation ,

1 mol of Ba(NO₃)₂ upon reaction with K₂SO₄ produces 1 mol of BaSO₄

                             OR

1 mmol of Ba(NO₃)₂ upon reaction with K₂SO₄ produces 1 mmol of BaSO₄

13.02 mmol of Ba(NO₃)₂ upon reaction with K₂SO₄ produces 13.02 mmol of BaSO₄

Molar mass of BaSO₄ is = 137+32+(4x16) = 233 g mol ^-1

So mass of BaSO₄ is m = number of moles x molar mass

                                    = 13.02x10 ^-3 mol x 233 g mol ^-1

                                    = 3.033 g

the theoretical yield of BaSO₄ is 3.033 g