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Name: __________________
THE MOLAR VOLUME OF HYDROGEN
Introduction:
In this experiment, you will be conducting a reaction of magnesium metal with excess hydrochloric acid in
a eudiometer (gas measuring tube). The volume of hydrogen will be equalized to the room pressure and the
final volume recorded.
Since the hydrogen gas will be collected over water, the hydrogen will be mixed with water vapour. The
pressure of the dry hydrogen gas can be calculated by subtracting the water vapour pressure from the room
pressure. Water vapour pressures at various temperatures are in the table below:
C
15
16
17
18
19
20
21
22
Table of water vapour pressure
C
23
24
25
26
27
28
29
30
kPa
1.71
1.82
1.94
2.06
2.20
2.34
2.49
2.64
kPa
2.81
2.98
3.17
3.36
3.57
3.78
4.00
4.24
With the volume, pressure and temperature of dry hydrogen gas, you can calculate the volume at standard
temperature and pressure (STP), 100 kPa and 273.15 K. The mass of magnesium used can be converted to
moles of magnesium used. The volume of dry hydrogen at STP produced by one mole of magnesium can
then be calculated. This is also the molar volume of hydrogen since 1 mol of Mg produces 1 mol H 2.
Suppose 0.0326 g of magnesium produces 35.6 mL of hydrogen measured over water at 100.0 kPa. The
room temperature is 26.0C (299K). The temperature of the water is 19.0C. From the table, we obtain a
water vapour pressure of 2.20 kPa at 19.0 C. The water vapour pressure is subtracted from the room
pressure, using Dalton’s Law of Partial Pressure, to obtain the pressure of the hydrogen gas alone:
100.0 kPa – 2.20 kPa = 97.8 kPa
The volume of hydrogen is corrected to STP:
35.6mL 
Thus, 0.00134 mol Mg (0.0326 g

97.8kPa 273K

 31.4mL
101.3kPa 299 K
1mol
) produced 0.0314 L H2 or 31.4 mL H2 measured at STP.
24.3g
Safety Instructions:

Wear safety goggles, latex gloves, and lab coat all the time.

Hydrogen is flammable and so ensure that the room is well ventilated and there are no sparks or
open flames in the laboratory.

Hydrochloric acid is corrosive and therefore if you get acid on your skin or clothes or in your eyes
wash it with plenty of cool water and inform the teacher.
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Purpose: To determine the volume of hydrogen gas produced when a given quantity of
magnesium reacts with excess hydrochloric acid.
Pre-Lab:
/7 K/U
/10 I
/4C
1) What is present in the eudiometer (gas collection tube) above the liquid phase, once the reaction in
this experiment is complete? Explain your answer. (2 K/U)
Hydrogen gas mixed with water vapour is present in the eudiometer above the liquid phase, once
the reaction is complete in this experiment.
In this reaction of magnesium metal with excess hydrochloric acid, hydrogen gas is formed. It is
collected over water in the eudiometer tube and so hydrogen gas mixed with water vapour is
present in the eudiometer above the liquid phase.
2) What is the reaction that is taking place in this experiment? Indicate all reactants and products
with subscripts. (2 I)
In this experiment magnesium metal reacts with excess of hydrochloric acid to form magnesium
chloride and hydrogen gas.
Mg (s) + 2 HCl (aq)→ MgCl2 (aq) + H2 (g)
3) If the room temperature is 23.5 C, and the atmospheric pressure is 99.5 kPa, what mass of
magnesium would be used to produce 75.0 mL of hydrogen gas? Show all calculations clearly.
(4 K/U)
Given:
Atmospheric pressure is 99.5 kPa.
Volume of hydrogen gas is 75 mL at 23.5 C
Room ambient temperature of 23.5 C equals to 273 + 23.5 = 296.5 0 K.
Standard atmospheric pressure is 101.3 kpa.
The volume of hydrogen corrected to STP can be calculated using Ideal Gas Equation
PV = n RT or
P1V1
P2V2
----------- = ---------T1
T2
=
75 mL x 99.5KPa x 273K
-------------- ------101.3
296.5
= 67.83 mL = 0.06783L
1 mole of ideal gas occupies 22.4 L/mol hence 0.06783L ÷ 22.4 L/mol = 0.00303 moles of
hydrogen gas are produced.
From the reaction, 1 mole of Mg produces 1 mole of hydrogen gas.
To produce 0.00303 moles of hydrogen gas we need 0.00303 moles x 24.31 g/moles = 0.074 g of
magnesium
 0.074 g of magnesium will be used to produce 75.0 mL of hydrogen gas.
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4) What is the volume of water vapour if 65.0 mL of hydrogen gas are produced? Explain your
answer. (2 I)
Assuming temperature of water is 18 0C and corresponding water vapour pressure is 2.06 kPa and
room temperature is 24 0C
Using Ideal gas law equation
P1V1 = P2V2
T1
T2
P1 = 273 + 18 = 291 kPa and P2 = 273 + 24 = 297 kPa
Water vapour pressure is 2.06 kPa and so pressure of hydrogen gas alone is
100 kPa- 2.06 kPa = 97.94 kPa
We can determine the volume of water vapour if 65.0 mL of hydrogen gas are produced.
2.06 x V1 = 97.94 x 65
291
297
V1 = 3028 mL = 3.028 L
 The volume of water vapour if 65.0 mL of hydrogen gas are produced is 3.028 L
5) Explain what equalizing the pressure of the gas in the eudiometer means. (I K/U)
Equalizing the pressure of the gas in the eudiometer means the pressure of hydrogen gas inside the
eudiometer must be same as the external atmospheric pressure or in other words, the level of the
water inside the tube must be the same as the level of water outside the tube.
6) Why must the pressure in the eudiometer be equalized after the reaction is complete? Explain
your answer. (2 I)
The pressure in the eudiometer must be equalized after the reaction is complete as the water
vapour pressure exerted on the hydrogen gas will affect the volume of hydrogen collected.
7)
Why is copper wire a good choice for the cage to hold magnesium? (1 I)
Copper wire is a good choice for the cage as it does not react with hydrochloric acid.
8) What would the volume of hydrogen be from the electrolysis of 0.0600 mol of water? Show all of
your work. (3 I)
In the electrolysis of water the reaction involved is
2H2O ( l ) → 2H2 (g) + O2 (g)
0.0600 mol of water produces 0.0600 mol of hydrogen gas
 The volume of hydrogen gas would be 0.0600 mol x 2.0158 g /mol = 0.120 g = 0.12 mL
(Molar mass of hydrogen gas is 2.0158 g)
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Prepare a flowchart of the procedure and a data table for the experiment (4 C)
Data Table:
Atmospheric pressure
99.5 kPa
Mass of Mg metal
0.074 g
Volume of H2 gas
79.18 mL
Room temperature
240C
Water temperature
180C
Water vapor pressure
2.06 kPa
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Materials:
 Eudiometer
 Copper wire
 Ring stand
 Buret clamp
 600 mL beaker
 100 mL graduated cylinder






One-hole rubber stopper (size 00)
Battery jar
Thermometer
Barometer
Magnesium ribbon
6 M HCl
Procedure:
1) Obtain between 0.07 and 0.08g of magnesium ribbon. Record the mass in your data table.
2) Roll the piece of magnesium into a coil. Use fine copper wire to build a copper cage around the
magnesium. The cage should allow the acid to enter but it should not have large openings though
which small pieces of magnesium might fall. The copper cage must be small enough to fit into the
gas measuring tube. Leave a long piece of extra copper wire attached to the cage to act as a
handle.
3) Set up a ring stand and buret clamp, and fill a 600 mL beaker about two-thirds full of water.
4) Carefully pour 10 mL of 6 M HCl into the eudiometer. Slowly fill the tube completely with tap
water. The 6M acid is denser than water and so it will remain at the bottom of the eudiometer. The
tube should be completely willed with water so that when a one-hoe stopper is placed in the tube a
little water will be forced out. Any bubbles sticking to the walls of the tube may be dislodged by
tapping the tube gently.
5) Holding the wire handle, lower the copper cage containing the magnesium into the water to a
depth of about 5 cm. Place the one-hole stopper into the mouth of the tube so that the wire handle
is caught between the glass and the stopper (not in the hole of the stopper).
6) Cover the hole in the stopper with your finger and invert the tube. Place the inverted tube in the
beaker of water. Remove your finger, and clamp the tube in position using the buret clamp. The
hydrochloric acid will eventually reach the magnesium ribbon and a reaction will occur. There is
no danger that the acid will reach the stopper before you remove your finger.
7) When the magnesium ribbon has disappeared and the reaction has finished, wait for 2 min to allow
the gas in the tube to come to room temperature. Dislodge any gas bubbles that are sticking to the
walls of the tube by gently tapping the tube.
8) Cover the hole in the stopper with your finger and transfer the tube to a battery jar or a large
graduated cylinder filled with water. Remove your finger and adjust the tube so that the level of
the water inside the tube is the same as the level of the water outside the tube. When the two levels
are equal, the pressure on the water outside the tube (room pressure) equals the pressure on the
water inside the tube caused by the gases in the tube. Read the gas volume. Drain and clean the
gas measuring tube. Record the volume in your data table
9) Measure the room temperature and the temperature of the water in the battery jar. Use this latter
temperature to find the water vapour pressure. Read the room pressure from the classroom
barometer. Record the temperature and pressure in your data table.
Analysis:
1) Calculate the number of moles of magnesium used. Show all calculations. (2 A)
If 0.074 g of magnesium is used then number of moles of magnesium =
0.074 g ÷ 24.31(molar mass of magnesium) = 0.00304 moles.
2) Calculate the pressure exerted by the dry hydrogen gas. Show all calculations. (1 A)
Using Dalton’s law of partial pressure, 99.5 kPa – 2.06kPa = 97.44 kPa. (At 180C temperature,
the water vapour pressure is 2.06 kPa from the table.)
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3) Calculate the volume that the dry hydrogen would have occupied at STP. Show all calculations.
(2 A)
If the volume of hydrogen gas from the experiment is 79.18 mL then
79.18 mL x 97.44 KPa x 273K
-------------------- = 70.22 mL of dry hydrogen at STP
101.3 kPa
297
4) What is the volume of 1 mol of hydrogen (the molar volume) at STP as determined by this
experiment? (2 A)
The volume of 1 mol of hydrogen (the molar volume) at STP as determined by this experiment is
If the volume that the dry hydrogen would have occupied at STP 70.22 mL which is equal to
0.0702 L from 0.00304 moles of magnesium then
The volume of 1 mol of hydrogen (the molar volume) at STP as determined by this
experiment would be 0.0704 L ÷ 0.00304 L /moles = 23.09 L
5) Calculate the percent difference between your answer and the accepted value for the molar volume
of hydrogen gas at STP. (2 A)
Percent Difference = Experimental Value – Actual Value
--------------------------------------------- × 100
Actual Value
23.09 – 22.4
--------------- × 100 = 3.08 %
22.4
Marking Scheme for Lab:
10 marks Communication based on communication rubric and proper lab format
18 marks Application (2 data table, 9 analysis, 2 error analysis, 2 conclusion, 3 lab performance)