Question

40. What is the experimental yield (in g of precipitate) when 17.2 mL of a 0.503 M solution of barium hydroxide is combined with 15.4 mL of a 0.556 M solution of aluminum nitrate at a 76.9% yield?

Answer 1

3 Ba(OH)₂ (aq) + 2 Al(NO₃)₃ (aq) --------------> 2Al(OH)₃ (s) + 3 Ba(NO₃)₂ (aq)

Ba(OH)₂ = molarity x volume = 0.503 M x 17.2 mL = 8.6516 mmol

Al(NO₃)₃ = molarity x volume = 0.556 M x 15.4 mL = 8.5624 mmol

From the balanced equation, we can say that

For every 3 moles of Ba(OH)₂ , 2 moles of Al(NO₃)₃ is required.

Hence,

For every 8.6516 mmol moles of Ba(OH)2 , ? mmoles of Al(NO₃)₃ is required.

                   ? = (8.6516 mmol/3 mol ) x 2 moles of Al(NO₃)₃

                            = 5.7677 mmoles of Al(NO₃)₃ is required

But, we have 8.5624 mmol of Al(NO₃)_3.

Therefore, Al(NO₃)_3. is in excess.

Hence, Ba(OH)2 is the Limiting reagent.

Theoretical yield of precipitate Al(OH)3 is calculated based on Limiting reagent i.e. Ba(OH)2.

3 Ba(OH)₂ (aq) + 2 Al(NO₃)₃ (aq) --------------> 2Al(OH)₃ (s) + 3 Ba(NO₃)₂ (aq)

3 mol                                                             2 mol

8.6516 mmol                                                    ?

Then,

                      ? = (8.6516 mmol/3 mol ) x 2 moles of Al(OH)₃

                        = 5.7677 mmoles of Al(OH)₃

Hence,

mass of Al(OH)₃ = mmoles x molar mass of Al(OH)3 = 5.7677 mmoles x 78 g/mol = 450 mg = 0.45 g

But it is given that experimental yield = 76.9%

Therefore,

     experimental yield of precipitate Al(OH)3 = 0.45 g x 76.9%

                                                    = 0.35 g