Antacid Stoichiometry with Gas Collected over Water
A student analyzes a 2.50 g sample of an antacid that contains calcium carbonate and inert ingredients. The student adds 50.0 mL of 1.00 M HCl, collects the carbon dioxide over water, and obtains 365 mL of gas at 22.0 °C and a barometric pressure of 755 mmHg. The vapor pressure of water at 22.0 °C is 19.8 mmHg. The student assumes the gas is dry CO2 and calculates the percent calcium carbonate without correcting for water vapor. The teacher notes that some CO2 remains dissolved in the aqueous solution and that the reaction may not have gone to completion during the 3 min collection period. The class must decide whether HCl or CaCO3 is limiting, calculate a corrected mass percent of CaCO3, and explain how each stated assumption would change the result.
1. Calculate the moles of CO2 corrected for water vapor. Determine which reactant is limiting and calculate the mass percent of CaCO3 in the antacid. Show all steps.
Read the worked answer
Correct CO2 pressure: P_CO2 = 755 − 19.8 = 735.2 mmHg = 735.2/760 = 0.9674 atm. T = 22.0 + 273.15 = 295.15 K; V = 0.365 L. n_CO2 = PV/RT = (0.9674 × 0.365)/(0.08206 × 295.15) = 0.01458 mol. Reaction: CaCO3 + 2HCl -> CaCl2 + CO2 + H2O, so n_CaCO3 = 0.01458 mol. Mass CaCO3 = 0.01458 mol × 100.09 g/mol = 1.459 g. Percent = (1.459/2.50) × 100 = 58.4%. HCl initially = 0.0500 L × 1.00 M = 0.0500 mol; HCl consumed = 2 × 0.01458 = 0.02916 mol, leaving 0.0208 mol. HCl is in excess, so CaCO3 is the limiting reactant.
2. Explain how each of the teacher’s two notes-dissolved CO2 and incomplete reaction-would affect the calculated mass percent. Also state whether ignoring water vapor makes the result too high or too low.
Read the worked answer
Ignoring water vapor makes the calculated CO2 pressure too high: using 755 mmHg instead of 735.2 mmHg gives n = 0.01497 mol and percent = 59.9%, so the result is too high by about 1.5 percentage points. Dissolved CO2 means some product is not measured as gas; the calculated n_CO2 is too low, so the mass percent is too low. Incomplete reaction also leaves some CaCO3 unreacted, again giving too little CO2 and a mass percent that is too low. The corrected 58.4% therefore still may be a lower bound if dissolved or unreacted CO2 is significant; the true percent could be higher. To improve, use a closed system or correct for solubility, swirl and wait longer, and confirm completion by testing for remaining carbonate or by adding excess acid.