- Level: International Baccalaureate
- Subject: Chemistry
- Word count: 912
Moles of Hydrogen report
Extracts from this document...
Introduction
Data Collection and Processing Raw Data Table 1- this table shows the results from our group's experiment Lengths of Magnesium Used Mass of Magnesium Used Moles of Magnesium Used Temperature (Kelvin) Patm PH2O VH2[dm3] 2cm 0.043 1.06483 x 1021 291.1K 102.5kPa 2.1kPa 0.036dm3 3.5cm 0.076 1.88203 x 1021 291.1K 102.5kPa 2.1kPa 0.0645dm3 Uncertainties * Length- �0.05cm * Volume- �0.1cm3 * Mass- �0.01g 1. Calculate the mass and number of moles of magnesium used in your experiment. Length of Magnesium- 2cm Moles of Magnesium Used = mass/ molecular mass = 0.043 / 24.31 = 0.0017689 Length of Magnesium- 3.5cm Moles of Magnesium Used = mass/ molecular mass = 0.076 / 24.31 = 0.0031263 2. From the partial pressure of water supplied calculate the partial pressure of hydrogen using the formula. P Atmosphere = P Hydrogen + P Water 102.5kPa = P Hydrogen + 2.1kPa P Hydrogen = 102.5kPa - 2.1kPa P Hydrogen = 100.4kPa Processed Data Table 2- this table shows the results of the class data Lengths of Magnesium (cm) ...read more.
Middle
/ 291.1 = (101.3X) / 273 3690 / 291.1 = (101.3X) / 273 12.676 = (101.3X) / 273 12.676 x 273 = 101.3X 3460.563 / 101.3 = X X = 34.162dm3 Length of Magnesium Used- 3.5cm (P1 x V1)/T1 = (P2 x V2)/T2 P1= 102.5kPa P2= 101.3kPa V1= 64.5cm3 V2= X T1= 291.1K T2= 273K (102.5 x 64.5) / 291.1 = (101.3X) / 273 6611.25 / 291.1 = (101.3X) / 273 22.711 = (101.3X) / 273 22.711 x 273 = 101.3X 6200.176 / 101.3 = X X = 61.206dm3 4. 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 + HCl --------> MgCl + H2 Mg + 2HCl --------> MgCl2 + H2 Mg- 1.501 H2 / Mg = X / 1.501 = 1 / 1 X = 1.501 Therefore, there are 1.501 moles of Hydrogen, when 1.501 moles of Magnesium was used. ...read more.
Conclusion
Analysis Section 3. From the graph determine the volume of hydrogen when the number of moles of hydrogen produced is 1.2 x 10-3 mol. Using the equation from question 5, y - x = 24.3 Substitute, 1.2 x 10-3, into the equation as x y - 1.2 x 10-3 = 24.3 y = 24.3 + 1.2 x 10-3 y = 24.3012 Therefore, the volume of hydrogen when the number of moles of hydrogen produced is 1.2 x 10-3 4. From your graph calculate a constant of proportionality between the two variables (gradient). Gradient = (Y2 - Y1) / (X2 - X1) = (67.8 - 43.5) / (2.789 - 1.789) = 24.3 / 1 = 24.3 5. Determine the equation that relates the variables using the constant of proportionality determine in 4. y= m x + b Because there is no y- intercept, therefore y = m x y = 24.3 x y - x = 24.3 ?? ?? ?? ?? The Molar Volume of Hydrogen Chemistry ...read more.
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