Chemistry dealing with yields

Question

Attempting to make barium phosphate by precipitation when mixing solutions of barium chloride and trisodium phosphate:

3 BaCl₂ (aq) + 2 Na₃PO₄ (aq) → Ba₃(PO₄)₂ (s) + 6 NaCl(aq)

208.33 163.94 601.92

0.461 g of BaCl₂ and 0.581 g of Na₃PO₄ were each dissolved separately in -50 mL H₂O in beakers. The Na₃PO₄ solution was poured into the BaCl₂ solution with mixing and the resulting suspension of white powder was digested at 80°C for 30 min to give 0.748 g of white solid after filtration, washing, and air drying.

1. Use gravimetric factor to calculate theoretical yield.

X-ray diffraction by the powder identified it not as barium phosphate, but as the mineral phase nabaphite:

BaCl₂(aq) + Na₃PO₄ (aq) ----> NaBaPO₄ - 9 H₂O (s) + 2NaCl(aq)

1. What is the theoretical yield of the isolated product and the observed percent yield?

When the same procedure was conducted, but digestion was carried out in a sealed Teflon bomb at 105°C for 3 hours instead of 80°C for 30 min, 0.352 g of white solid product was collected and identified by X-ray diffraction as Ba₃(PO₄)₂, not nabaphite.

What was the % observed yield of this product?

Answer 1

To calculate the theoretical yield of Ba₃(PO₄)₂, we will use the mole ratios of the balanced equation (stoichiometry) and the given masses of reactants. Since we are given masses of both reactants, we must find which, if either, is limiting. An easy way to do this is to simply divide moles of each reactant by the corresponding coefficient in the balanced equation. The lower value represents the limiting reactant...

For BaCl₂: 0.451 g x 1 mol / 208.33 g = 2.16x10^-3 moles (÷3->0.72x10^-3) = LIMITING

For Na₃PO₄: 0.581 g x 1 mol / 163.94 g = 3.54x10^-3 moles (÷2->1.77x10^-3)

Since BaCl₂ is limiting, we will use the moles (2.16x10^-3) to find the theoretical yield of Ba₃(PO₄)₂...

2.16x10^-3 mols BaCl₂ x 1 mol Ba₃(PO₄)₂ / 3 mols BaCl₂ = 7.20x10^-4 mols Ba₃(PO₄)₂

7.20x10^-4 mols Ba₃(PO₄)₂ x 601.92 g Ba₃(PO₄)₂ / mol = 0.433 g Ba₃(PO₄)₂ =Theoretical yield

The theoretical yield of nabaphite is still dependent on the fact the BaCl₂ is limiting. Thus...

2.16x10^-3 mols BaCl₂ x 1 mol NaBaPO₄ • 9 H₂O / mol BaCl₂ = 2.16x10^-3 mols NaBaPO₄ • 9 H₂O

2.16x10^-3 mols NaBaPO₄ • 9 H₂O x 417.43 g / mol = 0.902 g NaBaPO₄ • 9 H₂O = Theoretical yield

Observed % yield = actual yield / theoretical yield (x100%)...

Observed % yield of NaBaPO₄ • 9 H₂O = 0.748 g / 0.902 g (x100) = 82.9% yield

Observed % yield of Ba₃(PO₄)₂ = 0.352 g / 0.433 g (x100) = 81.3% yield