Question

Trial 1: P_H2O = 18.65mmHg; Trial 2: P_H2O = 18.65mmHg

(b) Using Dalton’s Law of partial pressures, calculate the pressure of hydrogen gas inside the eudiometer (in mmHg) for each trial. Assume that water and hydrogen are the only gases inside the eudiometer. (Hint: This is NOT a PV=nRT calculation.)

Trial 1:  

Trial 2:

17. Using your answer from Q3(b), calculate the mole fraction of hydrogen in the eudiometer for both trials.

Trial 1:                                                                        Trial 2:

17. Calculate the actual yield of hydrogen gas (in moles) produced in each trial. You may assume that the temperature of the hydrogen gas is equal to the temperature of the water bath. (use the ideal gas law.)

Trial 1:

Trial 2:

17. Calculate the percent yield (based on moles of hydrogen gas) for each trial.

Trial 1:                                                            Trial 2:

Data table template:

	Trial 1	Trial 2
Atmospheric pressure (mmHg)	745.9 mm Hg	746.3 mm Hg
Mass of Mg (g)	.0354 g	.0340 g
Volume of gas (mL)	36.05mL	34.10 mL
Water temperature (ºC)	21.0ºC	21.0ºC

Answer 1

Sol.

(1)

Trial 1 :

Partial pressure of hydrogen gas = P  

= Atmospheric pressure - Vapour pressure of water at 21°C

= 745.9 mm Hg - 18.65 mm Hg

= 727.25 mm Hg

Trial 2 :

Partial pressure of hydrogen gas = P  

= Atmospheric pressure - Vapour pressure of water at 21°C

= 746.3  mm Hg - 18.65 mm Hg

= 727.65  mm Hg

(2) Trial 1 :

Mole fraction of hydrogen gas = Partial pressure of hydrogen gas / Atmospheric pressure  

= 727.25 mm Hg / 745.9 mm Hg  

= 0.975

Trial 2 :

Mole fraction of hydrogen gas = Partial pressure of hydrogen gas / Atmospheric pressure  

= 727.65 mm Hg / 746.3 mm Hg  

= 0.975  

(3) Trial 1 :

Temperature = T = 21 °C = 21 + 273.15 K = 294.15 K

Volume of hydrogen gas = V = 36.05 mL = 36.05 / 1000 L = 0.03605 L

Gas constant = R = 62.363 L mm Hg / K mol

So , actual yield of hydrogen gas = n

= PV / RT

= 727.25 mm Hg × 0.03605 L / ( 62.363 L mm Hg / K mol × 294.15 K )

= 0.00143 mol  

Trial 2 :

Temperature = T = 21 °C = 21 + 273.15 K = 294.15 K

Volume of hydrogen gas = V = 34.10 mL = 34.10 / 1000 L = 0.03410 L

Gas constant = R = 62.363 L mm Hg / K mol

So , actual yield of hydrogen gas = n

= PV / RT

= 727.65 mm Hg × 0.03410 L / ( 62.363 L mm Hg / K mol × 294.15 K )

= 0.00135 mol  

(4)

Trial 1 :

Moles of Mg = mass of Mg / molar mass of Mg

= 0.0354 g / 24.305 g/mol = 0.00146 mol  

Theoretical yield of hydrogen gas = moles of Mg = 0.00146 mol  

Percent yield of hydrogen gas  

= ( actual yield of hydrogen gas / theoretical yield of hydrogen gas ) × 100

= ( 0.00143 mol / 0.00146 mol ) × 100

= 97.94  %

Trial 2 :

Moles of Mg = mass of Mg / molar mass of Mg

= 0.0340  g / 24.305 g/mol = 0.00140 mol  

Theoretical yield of hydrogen gas = moles of Mg = 0.00140 mol  

Percent yield of hydrogen gas  

= ( actual yield of hydrogen gas / theoretical yield of hydrogen gas ) × 100

= ( 0.00135 mol / 0.00140 mol ) × 100

= 96.43  %