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# Molar Volume of Hydrogen

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Introduction

Data Collection Table 1.0 - Quantitative data of the total mass and moles of a 3cm piece of magnesium, including temperature, pressure of atmosphere and the final volume of hydrogen created after the reaction. Moles of Mg used ( mol ) + 0.001 g Mass Mg used ( g ) + 0.001 g Volume of H2 (cm3) + 0.1 cm3 Temperature ( K ) + 0.1 K Pressure atm. (kPa) + 0.1 kPa Pressure H2 (kPa) + 0.2 kPa Pressure H2O (kPa) + 0.1 kPa 0.003 0.064 49.0 292.3 102.6 100.5 2.1 Table 1.1 - Table to show the length, corresponding mass of magnesium and the volume of hydrogen collected on reaction with hydrochloric acid, the average volume of hydrogen created and the moles of hydrogen gas created. 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 of H2 (SLC) (cm3) Mole of H2 used Mol �0.001mol 1.0 0.021 17.2 18.0 17.8 17.7 � 0.5 0.001 1.5 0.032 27.0 26.8 28.3 27.3�1.0 0.001 2.0 0.043 36.0 36.0 ...read more.

Middle

4. 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 5. 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 Observations 1. When the magnesium was attached to the copper and put into the hydrochloric acid there were no bubbles present at first, however, when the tube was flipped upside down tiny bubbles could be seen floating up. (less bubbles) 2. At the start of the experiment, only small bubbles could be seen closer to the magnesium than anywhere else 3. As soon as the acid reached the bottom of the flipped tube there were more bubbles present and lots of tiny bubbles filled the tube 4. ...read more.

Conclusion

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 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 + 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 ?? ?? ?? ?? The Molar Volume of Hydrogen The Molar Volume of Hydrogen . Jessica Huynh HL Chemistry - Mr Grant ...read more.

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