Ideal Gas Law: Stoichiometry Objectives: 1. To measure the mass of gas produced

Question

Ideal Gas Law: Stoichiometry Objectives: 1. To measure the mass of gas produced using temperature, pressure, and volume, 2. To calculate the theoretical yield of gas from starting quantities. 3 To determine the percent yield of gas. 4. To understand partial pressures and mole fractions of gases. Hydrochloric acid is corrosive. Avoid contact, and wash thoroughly with water after any contact eg procedure u f Weigh a short strip of magnesium ribbon on an analytical balance. If the mass of the ribbon is greater than 0.040 g. shorten the ribbon to reduce its mass to 0.040 g or less. Record the mass. 2 Use a thistle tube to add 15 mL of 3 M hydrochloric acid to a eudiometertutee 3. While holding the eudiometer tube at an angle to min mize the miina of he acid at the bottom, add water slowly within deionized a centimeter of the end. 4 Bend the weighed magnesium strip and push it into the open end of the eudiometer tube (like a small spring) so that it wil not fall out when the tube is invented 5 faing the eudiometer tube completely with water so no air bubble wil be trapped when the tube is inverted. Also add at approximately 350 mL of tap water to a 400 mL beaker. 6. Using your index finger to fully cover the eudiometer opening, invert the tube into the beaker, your finger when the tube opening is under water, If Remove the ribbon falls out of the tube, seek assistance from the instructor Hold the eudiometer tube upright (either by hand or with a ring stand) until the magnesium ribbon reacts. from magnesium stays at the bottom of the tube, the reaction will run faster if the tube is tited at about 30 degrees vertical. If the magnesium ribbon floats inside the tube, it is probably better to hold the tube straight up. When the reaction is complete, cap the eudiometer tube with your index finger and transfer the tube to the tall cylinder of room temperature Read and record the volume of gas in the tube when the water levels are the eudiometer same inside and outside the tube. Notice the scale increases going down the cudiometer tube. 9 Repeat steps 1-8 for the second trial. 10, obtain the barometric pressure in the room and record it 11. Measure the temperature of the water inthe tall cylinder and look of the vapor pressure of water for that temperature. Assume this is the temperature of the gas at the volume and pressure measured. Trial 1 Trial 2 Mass of magnesium ribbon Volume of gas colected 32.g Temperature of the gas 21.2 C, 20. 24, 12, t4 I

Explanation / Answer

Trial 1

Trail 2

Mass of magnesium ribbon

0.0389 g

0.0392 g

Volume of gas collected

32.8 mL

43.4 mL

Temperature of the gas

21.2C

20.8C

Barometric Pressure

29.12 mm Hg

29.12 mm Hg

Vapor Pressure of Water

18.88 torr

18.42 torr

Partial Pressure of Hydrogen gas

10.24 mm Hg (1 mm Hg = 1 torr)

10.70 mm Hg

A1) The balanced chemical equation for the reaction is

Mg + 2 HCl -----> MgCl2 + H2

As per the balanced stoichiometric equation,

1 mole Mg = 1 mole H2.

Mass of Mg ribbon taken = 0.0389 g. Molar mass of Mg = 24.305 g/mol.

Therefore, moles of Mg taken = (0.0389 g)/(24.305 g/mol) = 0.00160 mole.

Moles of H2 produced = (0.00160 mole Mg)*(1 mole H2/1 mole Mg) = 0.00160 mole H2.

Molar mass of H2 = 2.016 g/mol; therefore, mass of H2 produced = (0.00160 mole)*(2.016 g/mol) = 0.0032256 g 0.00322 g (ans).

A2) The balanced chemical equation for the reaction is

Mg + 2 HCl -----> MgCl2 + H2

As per the balanced stoichiometric equation,

1 mole Mg = 1 mole H2.

Mass of Mg ribbon taken = 0.0392 g. Molar mass of Mg = 24.305 g/mol.

Therefore, moles of Mg taken = (0.0392 g)/(24.305 g/mol) = 0.00161 mole.

Moles of H2 produced = (0.00161 mole Mg)*(1 mole H2/1 mole Mg) = 0.00161 mole H2.

Molar mass of H2 = 2.016 g/mol; therefore, mass of H2 produced = (0.0016 mole)*(2.016 g/mol) = 0.00324576 g 0.00324 g (ans).

B1) We need to convert the values of pressure, volume and temperature to match the unit of the gas constant, R.

Pressure of dry hydrogen = 10.24 mm Hg = (10.24 mm Hg)*(1 atm/760 mm Hg) = 0.01347 atm (1 atm = 760 mm Hg)

Volume of gas = 32.8 mL = (32.8 mL)*(1 L/1000 mL) = 0.0328 L.

Temperature of gas = 21.2C = (21.2 + 273) K = 294.2 K.

Use the ideal gas law: n = P*V/RT where the symbols denote pressure, volume, temperature and R = 0.082 L-atm/mol.K is the gas constant. Plug in values:

n = (0.01347 atm)*(0.0328 L)/(0.082 L-atm/mol.K).(294.2 K) = 1.831*10-5 mole.

Mass of hydrogen produced = (1.831*10-5 mole)*(2.016 g/mol) = 3.691*10-5 g (ans).

B2) Pressure of dry hydrogen = 10.70 mm Hg = (10.70 mm Hg)*(1 atm/760 mm Hg) = 0.01408 atm (1 atm = 760 mm Hg)

Volume of gas = 43.4 mL = (43.4 mL)*(1 L/1000 mL) = 0.0434 L.

Temperature of gas = 20.8C = (20.8 + 273) K = 293.8 K.

Use the ideal gas law: n = P*V/RT where the symbols denote pressure, volume, temperature and R = 0.082 L-atm/mol.K is the gas constant. Plug in values:

n = (0.01408 atm)*(0.0434 L)/(0.082 L-atm/mol.K).(293.8 K) 2.536*10-5 mole.

Mass of hydrogen produced = (2.536*10-5 mole)*(2.016 g/mol) = 5.112*10-5 g (ans).

C) Percent yield = (actual yield)/(theoretical yield)*100

For trial 1, percent yield = (3.691*10-5 g)/(0.00322 g)*100 = 1.146 (ans).

For trial 2, percent yield = (5.112*10-5 g)/(0.00324 g)*100 = 1.57777 1.578 (ans).

Trial 1

Trail 2

Mass of magnesium ribbon

0.0389 g

0.0392 g

Volume of gas collected

32.8 mL

43.4 mL

Temperature of the gas

21.2C

20.8C

Barometric Pressure

29.12 mm Hg

29.12 mm Hg

Vapor Pressure of Water

18.88 torr

18.42 torr

Partial Pressure of Hydrogen gas

10.24 mm Hg (1 mm Hg = 1 torr)

10.70 mm Hg

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