Molar Volume of a Gas Lab

Molar Volume of a Gas: Experiment Introduction -Objectives: The objectives of this experiment are to determine the molar volume of H 2 gas produced by the reaction: Mg (s) + 2 HCl (aq)  H2 (g)+ MgCl2 (aq), as well as calculate the molar volume of H2 gas at STP. -Theory: According to Avogadro, equal volumes of all gases at the same temperature contain the same number of particles regardless of the identity of the gas. His hypothesis states as well that at constant temperature and pressure, the volume of any gas is directly proportional to the number of moles of any gas. The molar volume is the volume occupied by one mole. The molar volume is defined by following formula: (V= molar volume, v =volume, n= moles) V= v/ mol The molar volume is being compared it to an ideal gas at standard conditions (0°C, 101.3 kPa). Molar volume is calculated through the following formula where V is molar volume, R is universal gas constant, T is temperature (K) and P is pressure (kPa). RT/P Dalton’s law of partial pressures is used to determine the partial pressure of hydrogen gas and the large graduated cylinder. The pressure punching gases collected as measured through water displacement. The formula for the partial pressure of hydrogen gas is the following: P= P H2 + P H2O Procedure Refer to the laboratory manual Data and Results Mass of Mg (g) 0.0323 g Moles of Mg (mol) 0.00133 moles Volume H2 gas (L) 34.0 mL Moles H2 gas (mol) 0.00133 moles Molar volume of H2 at Dawson’s temperature 25.6 L/mol Molar volume of H2 at STP (L) and pressure (L) 23.5 L/mol % error 4.78% Sample Calculations: 1. The number of moles of Mg 0.0323 g x (1 mol/ 24.31 g of Mg) = 0.00133 moles of Mg 2. The number of moles of H2 1 mole of H2 = 1 mole of Mg (1:1 ratio)  0.00133 moles of H2 3. The molar volume of H2 produced 0.034 L/ 0.00133 mol = 25.56 L/mol 4. The partial pressure of H2 P total = 772 mm Hg x 101.3 KPa / 760 mm Hg = 102.9 KPa P total = P H2 + P H2O P H2 = P total - P H2O = 102.9 KPa – 2.57 KPa = 100.33 KPa 5. The Molar volume of H2 and STP P1V1 / T 1 = P2V2 /T2 (100.33 KPa)(25.56 L) / (21.5°C + 273) = (101.3 KPa)(V)/ 273 V= 23.47 L/mol 6. Percent Error (Theoretical – Experimental / Theoretical) x 100 [(22.4 L/mol - 23.47 L/mol) / 22.4 L/mol ] x 100 = 4.78 % Discussion/conclusions After completing the experiment, as well as doing the calculations, the results obtained for the molar volume of hydrogen gas at STP was 23.47 L/mol. Therefore, the percent error calculated was 4.78% as the molar volume at STP is supposed to be 22.4 L/mol. This percent error is caused by sources of errors that occurred throughout the lab. Firstly, there may have been systematic sources of error, meaning that there were some inaccuracies of the measuring tools, such as the thermometer, the barometer, the beaker, the balance and the graduated cylinder. Secondly, it is possible that the conditions of pressure and temperature at Dawson/of the room changed causing for the percent error to increase. Overall, the objective was met as the goal of this experiment was to determine the molar volume of hydrogen gas at standard conditions. Even though there was the percent error of 4.78%, the objective was still completed. Reference General Chemistry Laboratory Experiments 202-NYA-05 (Revised Fall 2017). Dawson College. Post Laboratory Questions: 1. Assuming that the mass of magnesium is kept constant, what happens to the actual yield of the reaction if the molarity of the hydrochloric acid used is reduced when: a) The hydrochloric acid remains in excess The actual yield of the reaction will stay the same because HCl is not the limiting reagent, therefore, the values used in calculations will only acknowledge the mass of hydrochloric acid that is being reacted with, not its excess. b) The hydrochloric acid is the limiting reagent If the hydrochloric acid is the limiting reagent, the actual yield will change. It will decrease since the moles of hydrogen gas will decrease because not all of the magnesium was able to react. This will therefore decrease the final molar volume and result in a lower yield. 2. Explain how Avogadro’s hypothesis relates to the result obtained in this experiment Avogadro’s hypothesis states that at constant temperature and pressure, the volume of any gas is directly proportional to the number of moles of any gas. Therefore, the molar volume of the hydrogen gas at 237K and 1 atm should be equal to 22.4 L/mol. 3. In this experiment, you were asked to obtain the pressure in the gas collection tube through equalizing the water level inside the tube with the (external) water level in the large graduated cylinder. Why does this give you the pressure of the gas in the gas collection tube? By equalizing the water level of the gas collection tube with the external water level in large graduated cylinder, it is possible to measure the pressure of the hydrogen gas in the collection tube. When the two are aligned equally, the pressure inside the gas collection tube is equal to the atmosphere pressure. If we had not aligned them equally we would have to account for the water pressure within the cylinder in order to find the pressure of the hydrogen gas. 4. If the gas collection tube bubbles trapped in the water or on the string at the end of the experiment, explain the effect this would have on the reported value of the molar volume of the H2 gas? If the gas collection tube had bubbles trapped in the water or on the string at the end of the experiment, the value of the molar volume of the hydrogen gas will decrease because the volume of the gas contained within the bubbles was not accounted for. 5. A volume of 45.0 ml of gas at 27.0 and 758.0 torr, was produced when a Mg ribbon reacted with an excess of 6.0 M HCl a) What mass of Mg was used in the reaction? PV = nRT (101.033 kPa)(0.045 L) = n(8.31 kPaL/molK)(300K) n= 0.001824 moles (0.001824 moles of Mg)(24.31 g Mg/mol) = 0.044 g Mg b) What is the molar volume of H2 (L/mol) at the described conditions? (101.033 kPa)(V) = (1 mol)(8.31 kPaL/molK)(300K) v= 24.68 L/mol 6. In the reaction between HCl and Mg ribbon, 40.0 ml of a gas mixture, collected in a gas collection tube at constant T and P, contained H2 gas that was mixed with gaseous H2O. If 35.0 mL of the gas mixture contained 1.75 x 10-3 mole of H2 a) What mass of H2O was present in the gas mixture you collected? 40 mL – 35 mL = 5 mL H2O =0.005 L n1/v1 = n2/v2 0.00175 mol / 0.35 L = n2 / 0.005L n2 = 0.00025moles (0.00025moles x 18.02 g/mol) = 0.0045 g b) What is the mole % of H2O in the mixture? [ 0.00025moles / (0.00175 mol + 0.00025moles) ] x 100 = 12.5 %