Question

Date: 7 Record the barometric pressure. 8. Repeat this experiment with the second piece of magnesium ribbon. 9. Untwist the wire from the piece of fiberglass. Rinse the wire, fiberglass square, and stopper, place them in the container, and return the container to the instructor's desk. The table of temperature/Vapor pressure of water is in your textbook and in the back of your laboratory manual. The density of mercury is 13.6 g/mL. 61302083 Data Sheet Unknown Number Sample 2 40.8m2 Volume of H: gas (mL) 260 Temperature of water (°C) Sample 1 141.9. ML 23c 197mm 21.1 immag 40mmHg Height of water column (mm) 180mm 25.2 mm ita 740mm Hg Vapor pressure of water at above temperature (mmHg) Barometric pressure (mmHg) Calculations (Show calculations) Pressure of the column of water (mmHg) Pressure of dry hydrogen (mmHg) Moles of Hydrogen Mass of Magnesium ribbon (g) Average mass of magnesium ribbon (g) Temperature (°C) Pressure (mm Hg) 17 14.5 18 15.5 19 16.5 20 17.5 21 18. 7 22 1 9.8 23 21.1 Temperature (°C) ressure (mm Hg) 24 22.4 25 23. 8 26 25, 2 27 267 28 28, 3 29 30.0 30 31.8 Page 2

Answer 1

It is know that

mmHg *(density of Hg) = mmH₂O

For example, the pressure of the column of water in sample 1 is given as 197 mmH₂O. This pressure can be easily converted to mmHg as

mmHg *(13.6 g/mL) = 197 mmH₂O

=====> mmHg = (197 mmH₂O)/(13.6 g/mL)

=====> mmHg = 14.48

For sample 2, the pressure of the water column is given as 180 mm. The pressure in mmHg is given as

mmHg = (180 mmH₂O)/(13.6 g/mL)

= 13.23

Next fill up the table as below.

	Sample 1	Sample 2
Pressure of the column of water (mmHg)	14.48	13.23
Pressure of dry hydrogen (mmHg) = (barometric pressure) – (vapor pressure of water at the given temperature) – (pressure due to height of water column)	740 – 21.1 – 14.48 = 704.42	740 – 25.2 – 13.23 = 701.57
Moles of hydrogen (check calculation below)	0.001600	0.001536
Mass of magnesium ribbon (g) (check calculation below)	0.039	0.037
Average mass of magnesium ribbon (g)	½*(0.039 + 0.037) = 0.038

Moles of hydrogen

Sample 1:

Pressure of dry hydrogen = 704.42 mmHg

= (704.42 mmHg)*(1 atm)/(760 mmHg) = 0.9269 atm

Volume of hydrogen gas = 41.9 mL = (41.9 mL)*(1 L)/(1000 mL)

= 0.0419 L.

Temperature of hydrogen gas = 23ºC = (23 + 273) K = 296 K.

Use the ideal gas law.

P*V = n*R*T

where n is the number of moles of hydrogen gas.

Therefore,

n = PV/RT

= (0.9269 atm)*(0.0419 L)/(0.082 L-atm/mol.K)(296 K)

= 0.001600 mol.

Sample 2:

Pressure of dry hydrogen = 701.57 mmHg

= (701.57 mmHg)*(1 atm)/(760 mmHg) = 0.9231 atm

Volume of hydrogen gas = 40.8 mL = (40.8 mL)*(1 L)/(1000 mL)

= 0.0408 L.

Temperature of hydrogen gas = 26ºC = (26 + 273) K = 299 K.

Use the ideal gas law.

P*V = n*R*T

where n is the number of moles of hydrogen gas.

Therefore,

n = PV/RT

= (0.9231 atm)*(0.0408 L)/(0.082 L-atm/mol.K)(299 K)

= 0.001536 mol.

Mass of magnesium ribbon

The balanced stoichiometric equation is

Mg (s) + 2 HX (aq) ----------> MgX₂ (aq) + H₂ (g)

As per the stoichiometric equation,

1 mol H₂ = 1 mol Mg.

Atomic mass of Mg = 24.301 g/mol

Sample 1:

Mol(s) Mg = mol(s) H₂ = 0.001600 mol.

Mass of Mg = (0.001600 mol)*(24.305 g/mol)

= 0.038888 g

≈ 0.039 g

Sample 2:

Mol(s) Mg = mol(s) H₂ = 0.001536 mol.

Mass of Mg = (0.001536 mol)*(24.305 g/mol)

= 0.03733 g

≈ 0.037 g