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Molar Volume of a Gas Lab
General Chemistry (Dawson College)
Studocu is not sponsored or endorsed by any college or university
Molar Volume of a Gas Lab
General Chemistry (Dawson College)
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Molar Volume of a Gas: Experiment #3
Macey Zemel 1939367
Partner: Emma Rothstein
202-NYA-05, GENERAL CHEMISTRY, Section 00016, Jubrail Rahil
Tuesday, October 8, 2019
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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)
H
2
(g)+ MgCl
2
(aq), as well as
calculate the molar volume of H
2
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 H
2
+ P H
2
O
Procedure
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Refer to the laboratory manual
Data and Results
Mass of Mg (g)
0.0323 g
Moles of Mg (mol)
0.00133 moles
Volume H
2
gas (L)
34.0 mL
Moles H
2
gas (mol)
0.00133 moles
Molar volume of H
2
at Dawson’s
temperature
25.6 L/mol
Molar volume of H
2
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 H
2
1 mole of H
2
= 1 mole of Mg (1:1 ratio)
0.00133 moles of H
2
3.
The molar volume of H
2
produced
0.034 L/ 0.00133 mol = 25.56 L/mol
4.
The partial pressure of H
2
P
total = 772 mm Hg x 101.3 KPa / 760 mm Hg = 102.9 KPa
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P
total = P H
2
+ P H
2
O
P H
2
=
P
total - P H
2
O
= 102.9 KPa – 2.57 KPa
= 100.33 KPa
5.
The Molar volume of H
2
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.
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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.
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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 H
2
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 H
2
(L/mol) at the described conditions?
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(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 H
2
gas that was
mixed with gaseous H
2
O. If 35.0 mL of the gas mixture contained 1.75 x 10
-3
mole of H
2
a)
What mass of H
2
O was present in the gas mixture you collected?
40 mL – 35 mL = 5 mL H
2
O =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 H
2
O in the mixture?
[ 0.00025moles / (0.00175 mol + 0.00025moles) ] x 100 = 12.5 %
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