Limiting Reactant and Percent Yield in a Calcium Carbonate Precipitation
A student mixes 25.0 mL of 0.200 M CaCl2 solution with 30.0 mL of 0.150 M Na2CO3 solution. A white precipitate of CaCO3 forms. The mixture is filtered, and the precipitate is washed twice with 10 mL portions of deionized water and dried at 110 °C to constant mass. The recovered solid has a mass of 0.432 g. The student's laboratory partner claims the yield is close to 100% and that the supernatant should contain no calcium ions because calcium carbonate is insoluble. The student is not sure whether the precipitate is pure, whether the supernatant still contains excess reactant, or whether the drying step could have affected the result. The student needs to calculate the limiting reactant and theoretical yield, evaluate the percent yield, and decide what test would settle the question about excess calcium ions.
1. Determine the limiting reactant, the theoretical yield of CaCO3, and the percent yield.
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Calculate moles: n(Ca2+) = 0.0250 L × 0.200 mol/L = 5.00 × 10−3 mol. n(CO3^2−) = 0.0300 L × 0.150 mol/L = 4.50 × 10−3 mol. The reaction is 1:1, so carbonate is limiting. Theoretical CaCO3 = 4.50 × 10−3 mol × 100.09 g/mol = 0.450 g. Percent yield = (0.432 g / 0.450 g) × 100 = 96.0%. The partner is wrong that no calcium remains: Ca2+ is in excess by about 0.50 × 10−3 mol, although most of the calcium has precipitated.
2. Does the supernatant contain excess Ca2+? Describe a simple test and the expected observation.
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Yes. After filtering, the supernatant should contain excess Ca2+ because carbonate was limiting. Add a few drops of Na2CO3 solution to a portion of the filtrate. If Ca2+ is present, a white CaCO3 precipitate forms. A blank of deionized water should show no precipitate. Alternatively, ammonium oxalate gives a white calcium oxalate precipitate. This tests the supernatant, not the washed solid, so it directly checks whether calcium remains in solution.
3. The student obtains 0.468 g after drying at 110 °C. Offer two plausible explanations and say how to distinguish them.
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0.468 g is greater than the theoretical 0.450 g, so the solid cannot be pure dry CaCO3. Two plausible causes are: (1) the precipitate was not washed free of soluble salts such as NaCl from the supernatant, or it contains filter-paper fibers, adding mass; (2) the solid was not fully dried or the balance was misread. To distinguish, rewash the solid with several small portions of deionized water, dry again, and reweigh. A large drop suggests a soluble impurity; if the mass stays high, check for filter-paper contamination or weighing error. Testing the washings for chloride would confirm NaCl.