AIM: To measure the length of the Mg strip and the volume occupied by the hydrogen gas
produced by the chemical reaction: Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2 (g).
PROCEDURE: Obtain one piece of Mg ribbon, measure its length, and record the mass of 1 m of Mg ribbon.
Tie it to one end of a piece of thread. Measure about 10 ml of 2 M HCl (hydrochloric acid) and pour it into a
gas-measuring tube. Slowly fill the gas-measuring tube with water all the way to the top and push the piece
of Mg ribbon into the water at the top of the tube. Add about 200 ml of water at room temperature to a 400ml beaker. Set up a ring stand and clamp, and place the inverted gas-measuring tube into the water
beaker. After the magnesium has completely reacted, read the volume of the gas in the tube.
OBSERVATIONS AND DATA
a. Length of Mg ribbons
b. Mass of 1.0 meter of Mg
c. Volume of H2 gas in tube
d. Room temperature
e. Barometric pressure
f. Water vapor pressure at room
temperature
2.81 kPa =
101.3 kPa
4.1cm
0.74 g
30.6 mL
26 ºC
764 torr
21.1 torr
x _
760mmHg x = 21.081935 torr = water vapor pressure at room temp
CALCULATIONS
A. How many moles of Magnesium were used?
1. Convert the length of your Mg strip from cm to meters. Find its mass by comparing it to the mass of
1.0 meters of Mg.
*Put the length of
? meter
=
X grams
your Mg strip here
1-m length
mass of 1 meter
0.04 meter
1 meter
=
X grams
0.74
x
grams = 0.0296 grams
2. Calculate the number of moles of Mg in your strip from its mass. (This will be equal to the number of
moles of H2 gas you hopefully produced.)
0.00296 grams = 0.0012333 moles
24 g/mole
B. How many moles of hydrogen gas were produced?
1. Calculate the partial pressure exerted by the H2 gas in the tube: (use your reference tables to determine
the partial pressure of water vapor.)
Pb = P(H2) + P(H2O)
764 torr = x + 21.1 torr
x = 742.9 torr
2. Convert room temperature from Celsius to Kelvin:
Kelvinº = ºCelsius + 273
Kelvinº = 26 + 273
299 °K
3. Convert your measured volume of H2 (g) from units of mL to liters. Calculate the
equivalent volume of the H2 (g) at STP (use the COMBINED GAS LAW)
P(stp)V(stp)
=
P(lab)V(lab) Use this equation to determine the amount of H2 (g)
T(stp)
T(lab)
you collected in the tube, corrected to STP.
(760 mm Hg)x
273 K
=
(742.9 mmHg)(0.0306 L)
296 K
224960x = 22.73274 x 273
224960x = 6206.038
224960
224960
x = 0.0275873 L
4. Calculate the number of moles of H2 (g) you produced.
0.0275873 L = 0.0012316 moles
22.4 L/mole
C.
RESULTS AND THEIR ACCURACY
1. Using the chemical equation for the reaction you performed, what is the ratio of
magnesium used to hydrogen produced?
The ratio of magnesium used to hydrogen produced is 1:1.
2. Assume that the volume of hydrogen gas, corrected to STP, (step B 3 above), contains the exact same
number of moles as the amount of Mg you used. Using these results, calculate the volume of gas that
one mole would occupy.
Number of moles = 1 mole
Volume at STP
xL
0.0012316 moles
= 1 mole
0.0275873
xL
0.0012316 x L = 0.0275873
0.0012316
0.0012316
x L = 22.399562 liters
3. The accepted value for the molar volume of a gas is 22.4 liters. How does your
experimentally determined value compare with this accepted value? Calculate your percentage error.
The experimentally determined value is smaller than the accepted value.
% error = l absolute value – accepted value l x 100
accepted value
% error = 22.399562 – 22.4 x 100
22.4
= 0.000438 x 100
22.4
= 0.0019554%
CONCLUSIONS AND QUESTIONS
1. a. What is meant by the term “partial pressure of a gas”?
The pressure exerted by a separate gas in a mixture of gases.
b. State Dalton’s Law of Partial Pressures.
In a mixture of gases, the total pressure of the mixture is equal to the sum of the pressures that
each gas would exert by itself in the same volume.
2. Find the volume of the following masses of gases at STP:
a. 80 g O2
80 g
=
2.5 moles
2.5 moles x 22.4 L/mole = 56 Liters
32 g/mole
b. 10 g H2
10 g
=
2 g/mole
5 moles
5 moles x 22.4 L/mole = 112 Liters
c. 14 g N2
14 g
=
28 g/mole
0.5 moles
0.5 moles x 22.4 L/mole = 11.2 Liters
d. 66 g CO
66 g
=
28 g/mole
2.4 moles
2.4 moles x 22.4 L/mole = 53.76 Liters
3. How many liters would the following number of mole of an ideal gas occupy at
STP?
a. 0.25 moles
0.25 mole x 22.4 L/mole = 5.6 Liters
b. 0.50 moles
0.50 moles x 22.4 L/mole = 11.2 Liters
c. 1 mole
1 mole x 22.4 L/mole = 22.4 Liters
d. 2 moles
2 moles x 22.4 L/mole = 44.8 Liters
e. 2.5 moles
1.5 moles x 22.4 L/mole = 56 Liters
2. What happens to the other product of the reaction used in this experiment?
The other product of the reaction used dissolved into the water.