Determining the Molar Volume of a Gas Lab Report The purpose of the experiment is to determine the molar volume of hydrogen gas at standard pressure and temperature. Excess hydrochloric acid was placed in a eudiometer and decanted with deionized water. A piece of magnesium ribbon was trapped in a copper wire cage in the eudiometer in order to keep it in place as the more dense hydrochloric acid diffused downward in the inverted eudiometer. The eudiometer was placed in a water bath and the magnesium ribbon and hydrochloric acid reacted to form hydrogen gas. The rubber stopper was removed in order to balance the water levels in the eudiometer and the bath. The process was repeated. Chemical theories and concepts used in this experiment includes …show more content…
Vapor pressure (H2O): 17.444 torr = 17.444 mmHg
Correction: (263 mm * 1.00 mmHg) / 13.6 mmHg = 19.3 mmHg
Corrected Pressure of H2:
758 mmHg - 19.3 mmHg = Pressure of H2 + 17.444 mmHg
Pressure of H2 = 721 mmHg
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
(1.60 cm * 0.0200 g) / 1 cm = 0.0320 g
(0.0320 g * 1 mol * 1 mol H2) / (24.3050 g * 1 mol Mg) = 0.00132 mol H2
Volume at STP:
(760 mmHg)(V1)(292.9 K) = (721 mmHg)(.0314 L)(273 K)
V1 = 0.0278 L
Molar Volume: 0.0278 L / .00132 mol H2
= 21.1 L / mol
Trial 2
Pressure: (101.1 kPa * 760 mmHg) / 101.3 kPa = 758 mmHg
Vapor pressure (H2O): 17.444 torr = 17.444 mmHg
Correction: (268 mm * 1.00 mmHg) / 13.6 mmHg = 19.7 mmHg
Corrected Pressure of H2:
758 mmHg - 19.7 mmHg = Pressure of H2 + 17.444 mmHg
Pressure of H2 = 721 mmHg
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
(1.50 cm * 0.0200 g) / 1 cm = 0.0300 g
(0.0300 g * 1 mol * 1 mol H2) / (24.3050 g * 1 mol Mg) = 0.00123 mol H2
Volume at STP:
(760 mmHg)(V1)(292.9 K) = (721 mmHg)(.0302 L)(273 K)
V1 = 0.0267 L
Molar Volume: 0.0267 L / .00123 mol H2
= 21.7 L / mol
Average Molar Volume: (21.1 L/mol + 21.7 L/mol + 21.9 L/mol + 20.4 L/mol + 21.0 L/mol + 22.7 L/mol) / 6 = 21.5 L/mol
Percent difference: | 21.5 L/mol - 22.4 L/mol | / 24.4 L/mol = 0.0369 * 100% = 3.69%
Results Table