Limiting Reactant and Percent Yield in a Precipitation Reaction
A student mixes 50.0 mL of 0.200 M silver nitrate (AgNO3) with 50.0 mL of 0.150 M sodium chloride (NaCl) to produce silver chloride (AgCl) precipitate and sodium nitrate (NaNO3) in solution. The balanced equation is AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq). After filtering, drying, and weighing, the student obtains 1.02 g of AgCl. The student wants to know which reactant is limiting, the theoretical yield, and the percent yield. The molar masses are: AgNO3 = 169.87 g/mol, NaCl = 58.44 g/mol, AgCl = 143.32 g/mol.
1. Determine the limiting reactant and calculate the theoretical yield of AgCl in grams.
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Calculate moles of each reactant: Moles AgNO3 = 0.0500 L × 0.200 M = 0.0100 mol. Moles NaCl = 0.0500 L × 0.150 M = 0.00750 mol. From the 1:1 stoichiometry, AgNO3 requires 0.0100 mol NaCl, but only 0.00750 mol NaCl is available. So NaCl is limiting. Theoretical moles AgCl = moles NaCl = 0.00750 mol. Theoretical mass AgCl = 0.00750 mol × 143.32 g/mol = 1.0749 g ≈ 1.07 g.
2. Calculate the percent yield. Is the yield reasonable? What factors could cause a percent yield greater than 100%?
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Percent yield = (actual yield / theoretical yield) × 100 = (1.02 g / 1.0749 g) × 100 ≈ 94.9%. A yield of 94.9% is reasonable for a precipitation reaction. A yield greater than 100% could occur if the precipitate is not completely dry (water remains) or if impurities are present. Also, if the precipitate was not fully dried, the mass would be higher than expected.
3. If the student wanted to increase the percent yield, what experimental modifications could be made?
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To increase percent yield, ensure complete precipitation by allowing more time for settling, using a slight excess of NaCl (but not too much as it might cause other issues), washing the precipitate with a small amount of cold distilled water to remove soluble impurities without dissolving much AgCl, and drying thoroughly. Also, careful transfer of the precipitate from the filter paper can reduce loss.