Errors within Experiment:
Air bubbles in the gas measuring cylinder before the Magnesium was added.
Measuring the amount of Hydrochloric Acid
Measuring and cutting of Magnesium strips.
Magnesium strips broken into pieces.
Not taking the measurement of the volume of Hydrogen gas correctly due to parallax error.
Uncertainties within Experiment:
Measuring Cylinder ± 0.05mL
Gas measuring tube ± 0.1mL
Thermometer ± 0.05˚C
Barometer ± 0.05Kpa
Ruler ± 0.1mm
- Calculate the mass and number of moles of magnesium used in your experiment.
Number of moles used = mass of mg used (g)
gram formula mass of mg (g/mol)
= 0.021±0.1g
24.31
= 0.0008638
= 8.64 x 10 -4 ± 0.1g
8.64 x 10 -4 ± 0.1g moles of Magnesium was used.
Calculate the number of moles of Hydrochloric acid.
Moles of HCl = ( 8.64 x 10 -4) x 2
= 17.28 x 10 -4
= 1.73 x 10-3 ± 0.2
- From the partial pressure of water supplied calculate the partial pressure of Hydrogen using the formula.
P atmosphere = P Hydrogen+ P water
P atmosphere = 101.02 kPa ± 0.1Kpa
P Hydrogen =?
P water = 2.64 kPa
101.02 ± 0.1Kpa = Phydrogen + 2.64
Phydrogen = 101.02 ± 0.1Kpa – 2.64
Phydrogen = 98.38KPa ± 0.1Kpa
- The volume you measured under the laboratory conditions must be changed. Use the combined gas law to calculate the volume the hydrogen would occupy under STP conditions.
All gases occupy 22.4 dm3.
P1 = pressure – pressure of water = 99.8
P2 = 101.3Kpa
V1 = 17.7mL
V2 = V
T1 = 19.3 + 273 = 292.3K
T2 = 273K
P1V1 = P2V2
T1 T2
99.8 x 17.7mL = 101.3 x V
292.3 273
34.14 x 273 = 101.3V
1766.46 = 101.3v
101.3
V = 17.4mL
- Write a balanced equation for the reaction of magnesium with hydrochloric acid. From the equation determine the number of moles of hydrogen that can be formed from the moles of magnesium.
Equation - Mg + 2HCl = H2 + MgCl2
Mole ratio = 1: 2: 1: 1
Results:
- From your table construct a graph with moles of hydrogen (the variable you changed) on the x-axis and the volume of hydrogen (the variable you measured) on the y-axis.
Moles of hydrogen used against the volume of hydrogen produced.
Moles of hydrogen against the volume of hydrogen produced, with a trend line and equation of the graph.
- Describe the relationship between the two variables and indicate any proportionality that exists.
The amount of Hydrogen gas that was produced is generally proportional to that of the moles used in the reaction. Although one point on the graph that is not proportional, this is most likely due to an error.
Analysis Section:
-
From the graph determine the volume of hydrogen when the number of moles of hydrogen produced is 1.2 x 10-3 mol.
Approximately 25.0cm3 of Hydrogen gas would be produced when the moles of hydrogen is 1.2 x 10-3.
- From your graph calculate a constant of proportionality between the two variables (the gradient).
Points: A (0.00129, 27.4) B (0.00172, 36.3)
Gradient 1 = y2-y1
x2-x1
= 36.3 – 27.4
0.00172 – 0.00129
= 8.9
0.00043
= 20697.7cm3
= 20.7 dm3
Points: A (0.00344, 67.8) B (0.00387, 79.1)
Gradient2 = y2-y1
x2-x1
= 79.1 – 67.8
0.00387 – 0.00344
= 11.3
0.00043
= 26279.07 cm3
= 26.3dm3
Average = 10.1
0.00043
= 23488.4 cm3
= 23.5 dm3