Hydrogen Collection Over Water and an Impurity Claim
A general chemistry student reacts 0.245 g of magnesium ribbon with 50.0 mL of 2.00 M HCl in a eudiometer. The hydrogen gas is collected over water at 22.0°C. The atmospheric pressure is 744 mmHg, the vapor pressure of water at 22.0°C is 19.8 mmHg, and the measured gas volume is 248 mL. The student calculates a molar volume at STP of 21.1 L/mol and concludes that the magnesium was impure. The instructor asks her to check the limiting reactant, correct the gas pressure for water vapor, and decide whether the deviation is large enough to support the impurity claim. The reaction is Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g). Assume the gas is dry after correction and behaves ideally. The eudiometer was equilibrated to room temperature before reading, and no gas bubbles remained in the tube. Mg atomic mass 24.305 g/mol.
1. Identify the limiting reactant and calculate the theoretical moles of H2. How many liters of H2 at STP would that be?
Read the worked answer
Moles Mg = 0.245 g / 24.305 g/mol = 0.01008 mol. Moles HCl = 0.0500 L × 2.00 M = 0.100 mol. The balanced equation requires 2 mol HCl per mol Mg, so 0.01008 mol Mg needs 0.0202 mol HCl. HCl is in excess; Mg is limiting. Theoretical H2 = 0.01008 mol. At STP (273.15 K, 1 atm), V = n × 22.414 L/mol = 0.01008 × 22.414 = 0.226 L, or 226 mL.
2. Correct the measured gas pressure for water vapor and calculate the experimental molar volume at STP. Does the result support the student's impurity claim?
Read the worked answer
P_H2 = P_total − P_H2O = 744 − 19.8 = 724.2 mmHg = 724.2/760 = 0.953 atm. T = 22.0 + 273.15 = 295.15 K, V = 0.248 L. n_H2 from the gas = PV/RT = (0.953)(0.248)/(0.08206)(295.15) = 0.00976 mol. Convert the measured volume to STP: V_STP = 0.248 × (724.2/760) × (273.15/295.15) = 0.219 L. Molar volume = 0.219/0.00976 = 22.4 L/mol. If one instead divides by the theoretical 0.01008 mol Mg, the result is 21.7 L/mol, about 3% low. Either way, the deviation is small and consistent with gas loss or measurement error, not proof of impure magnesium. The student's 21.1 L/mol is too low to support the claim without further evidence.
3. If she had used the total pressure instead of the dry H2 pressure, would her calculated moles of H2 be too high or too low? Would that explain a falsely low molar volume?
Read the worked answer
Using P_total = 744 mmHg instead of P_dry = 724.2 mmHg makes the calculated n = PV/RT too high by the ratio 744/724.2 = 1.027, about 2.7%. Since molar volume is V_STP/n, an artificially high n makes the calculated molar volume too low. That direction matches the student's low result, but the correction alone changes 22.4 to about 21.8 L/mol, not 21.1. Additional small gas loss or volume/temperature error would be needed to reach 21.1. The vapor-pressure correction is required, but it does not by itself prove the Mg was impure.