Results Section
- 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)
Graph 1.0 - Shows the relationship between how the moles of hydrogen and how it effect the total volume of hydrogen.
- Describe the relationship between the two variables and indicate any proportionalities that exist.
From the graph we can see that as the moles of hydrogen increase, the volume of hydrogen is also increased.
Analysis Section
Results Section
- 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)
Graph 1.0 - Shows the relationship between how the moles of hydrogen and how it effect the total volume of hydrogen.
- Describe the relationship between the two variables and indicate any proportionalities that exist.
From the graph we can see that as the moles of hydrogen increase, the volume of hydrogen is also increased.
Analysis Section
-
From the graph determine the volume of hydrogen when the number of moles of hydrogen when the number of moles of hydrogen produced is 1.2x10-3 mol.
y = 16246.33x + 5.119
y = 16246.33 x (1.2 x 10-3) + 5.119
= 24.615 cm3
Therefore the volume of hydrogen is approximately 24 cm3.
- From your graph calculate a constant of proportionality between the two variables (the gradient).
Gradient = y2 – y1 / x2 - x 1
= 25.296 – 74.035 / 0.001 – 0.004
= 16246.33
- Determine the equation that relates that variables using the constant of proportionality determine in 4.
The equation of the line: y = mx + c
Where;
y = volume of hydrogen and x = moles of hydrogen
Therefore ;
y = 16246.33x + 5.119
Table 1.2 - Quantitative observation produced during the experiment
Extra Questions
- Calculator the mass and number of moles of magnesium used in your experiment.
- The mass of a 3 cm piece of magnesium is 0.0639g;
- The number of moles of magnesium used in the experiment
= mass / atomic mass x [ 6.02 x 1023 ]
= 0.0639/24.31 x [ 6.02 x 1023 ]
= 1.073 x 1021
-
Form the partial pressure of water supplied calculate the partial pressure of hydrogen using the formula ; Patmosphere = Phydrogen +Pwater
102.6 = Phydrogen + 2.1
Therefore Pressure of Hydroogen = 102.6 – 2.1 = 100.6 kPa + 0.2 kPa
Where;
- P Atmosphere = 102.6
- P Water = 2.1
- P Hydrogen =100.5
- 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 condition.
Combine gas law : P1V1 / T1 = P2V2 / T2
Where ;
P1 = 100.5 kPa
T1 = 292.3 K
V1 = 49.0 cm3
V2 = ?
Therefore the volume of hydrogen is;
100.5 x 49 / 292.3 = 101.3V / 273
16.8 = 101.3V / 273
4586.4 = 101.3V
V = 45.4 cm3 ±0.20
- 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 in each experiment.
Mg + 2HCl = H2 + MgCl2
Moles of magnesium = 0.003mol
Moles of hydrogen = 1 x 0.003 mol = 0.003 mol ± 0.001g
Where error of moles of magnesium = ± 0.001g