Impure Calcium Carbonate, Limiting Reagent and Gas Volume
In a school laboratory, 10.0 g of an impure calcium carbonate sample containing 80% CaCO3 by mass is treated with 100 mL of 2.0 M hydrochloric acid. The reaction is CaCO3 + 2HCl → CaCl2 + H2O + CO2. The molar mass of CaCO3 is 100 g mol⁻¹, and the molar volume of a gas at STP is 22.4 L mol⁻¹. A student predicts that the sample will produce 2.24 L of carbon dioxide at STP, because 10.0 g of CaCO3 is 0.10 mol. Another student says the acid may be limiting and the volume will be different. The mixture is warmed gently until no more gas is evolved. The class is asked to identify the limiting reagent, calculate the actual volume of CO2 at STP, and decide what would happen with only 50 mL of the same acid.
1. Calculate the moles of pure CaCO3 and of HCl. Identify the limiting reagent and calculate the volume of CO2 at STP.
Read the worked answer
Pure CaCO3 mass = 80% of 10.0 g = 8.0 g. Moles CaCO3 = 8.0/100 = 0.080 mol. Moles HCl = 0.100 L × 2.0 mol L⁻¹ = 0.200 mol. From the equation, 0.080 mol CaCO3 needs 2×0.080 = 0.160 mol HCl. HCl is in excess (0.040 mol left), so CaCO3 is limiting. Moles CO2 = 0.080 mol; volume at STP = 0.080×22.4 = 1.792 L ≈ 1.79 L.
2. Evaluate the first student's prediction of 2.24 L and explain the error.
Read the worked answer
The first student used the whole 10.0 g as if it were pure CaCO3: 10.0/100 = 0.10 mol, giving 0.10×22.4 = 2.24 L. But only 80% is CaCO3, so the actual amount is 0.080 mol. The prediction is therefore too high by about 0.45 L.
3. If only 50 mL of the same 2.0 M HCl is used with the same sample, identify the limiting reagent and calculate the CO2 volume at STP.
Read the worked answer
With 50 mL of 2.0 M HCl, moles HCl = 0.050×2.0 = 0.100 mol. 0.080 mol CaCO3 would need 0.160 mol HCl, but only 0.100 mol is present, so HCl is now limiting. Moles CO2 = 0.100/2 = 0.050 mol; volume = 0.050×22.4 = 1.12 L at STP.