CHM135 Lab 2 Report Sheet pdf

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Chemistry

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Dec 6, 2023

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Experiment 2: Gas Stoichiometry Report Sheet You are welcome to seek guidance on the content of this lab report from your TA during your lab session and from the lab coordinator during lab office hours. You are also welcome to work constructively with your peers on the general content and understanding of the material. However, the work submitted here in this report sheet must be your own. For more details on academic integrity and potential penalties see the Code of Behaviour on Academic Matters . I certify that this submitted laboratory report represents entirely my own efforts. I have read and understand the University of Toronto policies regarding, and sanctions for, academic honesty. I regret that I violated the Code of Behaviour on this assessment and would like to admit that now so that I can take responsibility for my mistake. Portions of this lab report will be auto-graded, so please follow the formatting in this document to ensure your results can be properly graded. 1. Write the balanced chemical equation for the reaction between HCl and Mg. Mg(s) + 2HCl(aq) = MgCl2 (aq) + H2 (g) 2. Did the pH change as the reaction occurred? Explain this observation. (Hint: What do you think is the limiting reactant?) No, the pH did not change. We measured pH with the indicator paper strip. Before and after the reaction it was a bright red colour, which corresponds to the pH of 1 (acidic), it tells us that HCl was not a limiting reactant. 3. Write the balanced chemical equation for the reaction that occurred when the hydrogen bubbles were lit on fire. 2H2 + O2 = 2H2O

4. Complete the table below using your experimental data. Some of the numbers in this table will be autograded and so include values only, do not include units, and if you would like to use scientific notation use exponential notation (e.g. 1.2 x 10 -5 is 1.2e-5 with no spaces between the "e" and the numbers). Temperature ( o C) Mass of Mg(s) (g) Volume of 1.0 M HCl (mL) Volume of H 2 (g) (mL) Moles of H 2 (g) (mol) Percent Yield (%) 25 0.0350 10,0 27 1.1e-3 76.67% Show your calculation for the number of moles of H 2 ( g ): PV=nRT, P=1 atm, V = 27ml (0.027L) , R = 0.08206 atmL/Kmol, T = 25 ° C (298 K) n = PV/RT n = [(1 atm)(0.027L)]/[0.08206 atmL/molK)(298K)] n = (0.027 atmL)/(24.45388atmL/mol) n = 0.00110, or 1.1e-3 moles of H2(g), (actual mass). Show your calculation for percent yield: According to the equation of the reaction, we know the mass of Magnesium, and the proportion of Mg to H2 (1:1), we can use this ratio to determine the theoretical amount of H2 for the yield equation. Mass of Mg(s) = 0.0350g Mg mm = 24.30 g/mol Molar mass = 0.0350g (1mol/24.30 g/mol) = 0.00144 mol Actual mass = 0.00110 mol Theoretical mass = 0.00144 mol Yield % = (actual/theoretical) (100%) = (0.00110/0.00144)(100%) = 76.67%

5. For the data collected by your lab section, what is the mean and standard deviation of the percent yields of hydrogen gas? The values in this table will be autograded and so include values only, do not include units, and if you would like to use scientific notation use exponential notation (e.g. 1.2 x 10 -5 is 1.2e-5 with no spaces between the "e" and the numbers). Note: Before you report your values below remember standard deviation is a measure of the spread in the data and so can tell you something about your confidence in the mean value and the number of appropriate significant figures. See the “After Lab 2” page on Quercus for more details. Mean (%) Standard Deviation (%) 83.72% 40.37% 6. Were some of the yields calculated in your lab section above 100%? This is not physically possible; however when we make these measurements, we are using volume as a measure of the number of moles of gas present. How might temperature confound the relationship between volume and moles of gas in this experimental setup? Yes, there were some results above 100%. We assumed the temperature of the reaction was the same as the temperature of the water in the water bin. This reaction is very sensitive to temperature change, and because of the inaccuracies in measurements results could be wrong. 7. In the second part of the experiment temperature is less of an issue as the reaction does not exchange as much heat with the solution. If we take temperature out of the equation, the biggest confounding factor is gas loss, either from leaks in the system or the trapping of gas in the tubing that was filled with water. If gas were lost during the experiment, would you expect the calculated atomic mass of the metal to be higher or lower than the actual atomic mass? Explain your answer. In the equation PV=nR, we are not considering the temperature. If the gas escaped in the reaction, we would notice a difference in the volume of gas in the cylinder (Lower V value). When the volume is decreasing, the moles’ concentration is decreasing. If we calculate the low number of molecules of the unknown mass, then atomic mass would be higher that it should be.

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