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Molar volume of hydrogen

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Introduction

Aim: To determine the volume of Hydrogen gas is produced when Magnesium reacts with Hydrochloric acid. Materials: Coil of Copper Wire, Hydrochloric Acid (3mol), 1.0 cm & 1.5 cm pieces of Magnesium strip. Stand and Clamp, Barometer, Thermometer, Beaker (200ml), Gas measuring tube Method: As on the experiment sheet Data Collection: Table 1.1 - The Length of the Magnesium strips with corresponding volumes of Hydrogen gas produced. Mg Length ( cm ) + 0.1 cm Mass Mg ( g ) + 0.001 g Volume 1 ( cm3 ) + 0.1 cm3 Volume 2 ( cm3 ) + 0.1 cm3 Volume 3 ( cm3 ) + 0.1 cm3 Average Volume ( cm3 ) + 0.3 cm3 1.0 0.021 17.2 18.0 17.8 17.7 1.5 0.032 27.0 26.8 28.3 27.4 2.0 0.043 36.0 36.0 37.0 36.3 2.5 0.053 49.0 43.5 46.3 3.0 0.064 56.0 53.8 54.9 3.5 0.076 58.4 58.4 4.0 0.085 67.8 67.8 4.5 0.094 77.0 81.2 79.1 Table 1.2 - Moles of hydrogen and the volume of hydrogen produced Moles of Hydrogen (mol) ...read more.

Middle

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 1. 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 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 =? ...read more.

Conclusion

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. 2. 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: 3. 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. 4. 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 ?? ?? ?? ?? Chemistry Chemistry ...read more.

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