CHEM San Diego State University Pressure of Dry Hydrogen Gas Essay

Post Lab Questions
I
J
,
1
· Why was it_ not necessary to discard and refill the generator flask with fresh 6 M HCI for subsequent reactions? Show a calculation to demonstrate your reasoning.
2.
In an experiment similar to yours, students were given an unknown sample that produced H 2S
or NH 3 gas. What problem(s), if any, might arise if the same procedures as this experiment were
followed?
3. Why is it not necessary to consider the amount of air that was in the generator flask at the start
of the reaction?
•
4.
0
Ifthere was a leak in the rubber tubing that allowed approximately 150 mL ofH 2 to escape, how
would it affect the relative amounts (% values) that you calculated as the results? Would the
%magnesium in the sample appear to be high, low, or remain the same? Show a calculation to
justify your answer.
–
Analysis of a Two-Component Alloy
5.
What assumptions do we make regarding the temperatures of the generator and collection
flasks? Are they valid?
6. How would the calculated number of moles of H2 collected change if the generator flask were
not given time to fully equilibrate?
7. The molecular mass of a gas can be determined using an application of the Ideal Gas Law:
::,V = nRT
and
m
n = —
MM
so
PV = …!!!.._R7
MM
(EQ 9.25)
An unknown gas with a mass of 1.545 ± 0.001 g at a temperature of 298.5 ± 0.1 K and a pressure of715 ± I mm Hg is found to occupy 0.260 ± 0.001 L. Determine the molar mass and
absolute error of this unknown. Assume that the gas constant, R, has no error.
, _ EXPERIMENT g
Analysis ofa TwoComponent Alloy
Background
In th.is experiment, you will react a weighed sample of a two-metal alloy with excess acid and collect the hydrogen gas evolved over water. If you measure the volume, temperature, and total pressure of gas and use the Ideal Gas Law, taking into account the pressure of water vapor in the
system, you can calculate the number of moles of hydrogen gas produced by the sample. The percent composition of the alloy will be found.
PV
= nRT
fl =
PV
RT
(EQ 9.1)
where Pi s the partial pressure of hydrogen gas. The volume, V. and the temperature T of the hydrogen are easily obtained from the data . The pressure of the d1y hydrogen, PH , requires more attention. Dalton ‘s Law says that the total pressure is equal to the sum of the partial pressure of the
individual gases in the mixtures:
P, u1al
=
LP,
(EQ 9.2)
So, the total pressure P har, is equal to the partial pressure of the hydrogen, PH , plus the partial
2
pressure of the water vapor, P 11 , 0 , plus or minus the water head pressure (as explained below), Poh·
For this experiment Phar is equal to the measured barometric pressure.
(EQ 9.3)
The water vapor in the bottle is present with liquid water so the gas is saturated with water vapor:
the pressure P1-1 0 under these conditions is equal to the vapor pressure of water at the temperature
of the experime:11. This value is constant at a given te mperature, and can be found in your textbook.
Once the water vapor pressure is known it can be subtracted from equation 9.3. Now, the pressure,
Pt>ar is equal to the pressure of the dry hydrogen and the head pressure.
The head pressure is measured by taking the difference in the heights of the liquid leve ls in the
large bottle and beaker (not the reaction fl ask). Thi s level represents the pressure differe nce in 111111
H, O between atmospheric pressure, P,.”,,, and the pressure inside the large botth:: and rcadic>n flask .
Analysis of a Two-Componen t Alloy
It is important to note that this difference must be converted to mmHg before it can be added to or
subtracted from the atmospheric pressure. Since the density of Hg is 13.6 times as dense as water,
the height difference values in mm must be divided by 13 .6 to convert them to mmHg.
For example if you measured a height difference of 320. mm between the bottle and the beaker, the
head pressure would be calculated as:
Pt..h
320. mm H 0
13.6
= – – – – 2- = 23.5
mm
H
g
(EQ 9.4)
If the water level in the beaker is higher than the level in the large bottle then the pressure in the
bottle is greater than atmospheric pressure. You would add the head pressure in Equation 9.3.
FIGURE 9.1
Gas Generation and Collection Setup: Beaker water level higher than bottle
dJlulc 11