Question

A chemist prepares a sample of chromium(II) phosphate by mixing together 100.0 mL of a 0.250 M CrCl₂(aq) solution with 150.0 mL of a 0.100 M Na₃PO₄(aq) solution. The chromium(II) phosphate precipitate formed is filtered off, dried, and its mass is 2.39 g. What is the percent yield of chromium(II) phosphate? (Use molar masses: CrCl₂ = 123; Na₃PO₄ = 164; chromium(II) phosphate = 346 g/mol).

Answer 1

The percent yield of chromium(II) phosphate is 83.0%, calculated by comparing the actual yield of 2.39 g to the theoretical yield of 2.88 g.

Step-by-step explanation:

Calculating Percent Yield for Chromium(II) Phosphate

To calculate the percent yield of chromium(II) phosphate, we must first determine the theoretical yield, which is the maximum amount of product that could be formed from the given reactants. In this reaction, chromium(II) chloride (CrCl₂) reacts with sodium phosphate (Na₃PO₄) to form chromium(II) phosphate (Cr₃(PO₄)₂) and sodium chloride (NaCl).

The balanced equation for this reaction is:
3 CrCl₂ + 2 Na₃PO₄ → Cr₃(PO₄)₂ + 6 NaCl

From the provided concentrations and volumes of reactants, we can calculate the moles of each reactant:

Using stoichiometry from the balanced equation, the limiting reactant is CrCl₂. Thus, the maximum number of moles of chromium(II) phosphate that can be formed is 0.0250 mol / 3 = 0.00833 mol of Cr₃(PO₄)₂.

The theoretical yield in grams of chromium(II) phosphate is:
0.00833 mol * 346 g/mol = 2.88 g

With an actual yield of 2.39 g, the percent yield is calculated as:

(Actual yield / Theoretical yield) * 100%
= (2.39 g / 2.88 g) * 100%
= 83.0%