Electrolysis splitting water

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Electrolysis splitting water ( electrolysis-splitting-water )

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One mole of an ideal gas takes up 22.4 L at room temperature and pressure. Q19. Calculate how many grams of chlorine gas were produced. First calculate how much hydrogen gas was produced. (________mL H ) 1L 1mol H2  = ________mol H gas produced 2 1000mL  22.4L  2 about 8x10-5 mol H2 Expected number of moles of Cl2 gas produced (use the expected ratio of H2:Cl2) : We expect the same amount of Cl2 to be produced as H2. Look up the molar mass of Cl in the periodic table Molar mass of Cl: _____35_______ g/mol Calculate mass of chlorine gas produced (remember to multiply by 2 to get g/mol of Cl2):  gCl (________ mol Cl2 )  _______ 2  = ________ g Cl2 produced should be about 0.006 g Q20. Calculate how much Chlorine would be needed to saturate the entire test-tube of water. Approximately how much water is in the test-tube? _____about 10 to 15______ mL Calculate mass of water in the test-tube:  1g   1kg  (________ mL)  1 mL    1000 g  = ________ kg water will be about 0.015 kg water Using the solubility chart in the back of your lab, how much chlorine would be required to saturate the entire test-tube?  molCl2    gCl  (________kgwater)_________ 2 solubility=________gCl2 tosaturatesolution  kgwater  Q21. Did you make enough Cl2 gas to saturate the solution? No. Q22. Why did you see so little gas at the anode (positive terminal)? All the chlorine gas was able to dissolve in the water Q23. In part 1, why did you see the oxygen and hydrogen gases actually produced – why didn't most of those dissolve into the water as well? Hint: look at the solubility charts provided. O2 and H2 are much less soluble in water (it takes a lot fewer grams of these gases to saturate the test-tube), so the gas that was produced remained mostly undissolved. Created by LABScI at Stanford 8

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