The Volume of One Mole of Gas Under Conditions of Room Temperature and Pressure

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The volume of one mole of gas under conditions of room temperature and pressure

Data collection and processing
Quantitative data in table with units and uncertainties
| Mass of Magnesium(Mg) (g)| Volume of Hydrogen (H₂) (cm³)| 1st trial| 0.040| 39.9|
2nd trial| 0.040| 40.3|
3rd trial| 0.035| 36.5|

Quantitative data:
The Magnesium was silvery-white, lustrous and relatively flexible before being placed in the burette. Whilst reacting with the hydrochloric acid, it dissolved and bubbles were visible. The hydrochloric acid slowly diffused downwards. Calculations:

Volume of un-diffused hydrochloric acid = 25 cm³
Average mass of magnesium = 0.038 g
Average volume of diffused hydrochloric acid = 11.1 cm³
Average volume of hydrogen produced = 38.9 cm³
| Mass of Mg(g)| Volume of hydrochloric acid (cm³)| Volume of water (cm³)| Volume of diffused hydrochloric acid(cm³)| Volume of hydrogen (cm³)| 1st trial| 0.040| 25| 50| 10.1| 39.9|

2nd trial| 0.040| 25| 50| 9.7| 40.3|
3rd trial| 0.035| 25| 50| 13.5| 36.5|
Average | 0.038| 25| 50| 11.1| 38.9|

Uncertainties:
Experimental error:
mass of Mg-average volume of H2 mass of mgX 100
= 24.00-38.9024.00X100
= -62.08 %
Equipment error: the burette used had an error of ±0.1 cm³ Therefore: 0.1 38.9X100= 0.26 %
Also, the balance used had an error of ±0.05
Therefore: 0.0500.038 X100= 131.58 %
Total equipment error = 0.26 + 131.58 = 131.84 %
Mole calculations:
Mg + 2HCl → MgCl₂ + H₂
Ratio: 1:2:1:1
Number of moles of Mg= 0.038 / 24.000
= 0.001 moles
Therefore 0.001 moles of H₂ occupies 38.9 cm ³

Conclusion and evaluation
The original equation for this experiment was Mg + 2HCl → MgCl₂ + H₂. The average volume of hydrogen produced was found to be 38.9 cm³. This was calculated by finding the difference of the un-diffused hydrochloric acid plus water from the diffused hydrochloric acid (50 cm³ -11.1 cm³). This result is in disagreement to the...
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