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Question 5 5 pts A pair of students collected the following data. What is the correct mole ratio based on the data collected?


Computer 9 MATERIALS Venier computer interface computer Temperature Probe two 10 mL graduated cylinders two 25 mL graduated c
10. Examine the graph to calculate and record the maximum temperature change. a. To determine the highest temperature, click
Determining the Mole Ratios in a Chemical Reaction DATA TABLE Volume S203 Temperature change Volume OcIF (mL) (mL) DATA ANALY
Question 5 5 pts A pair of students collected the following data. What is the correct mole ratio based on the data collected? Temperature CIO S203 Change 20 30 30 20 10 40 40 10 2.6 6.5 10.1 6.3 5 45 1.4
Computer 9 MATERIALS Venier computer interface computer Temperature Probe two 10 mL graduated cylinders two 25 mL graduated cylinders two 50 mL graduated cylinders three 250 mL beakers 0.50 M sodium hypochlorite, NaOCI, solution 0.50 M sodium thiosulfate, Na2S203, solution in 0.2 M sodium hydroxide, NaOH Styrofoam® cups PROCEDURE 1. Obtain and wear goggles. 2. Connect a Temperature Probe to Channel 1 of the Venier computer interface. Connect the interface to the computer with the proper cable. 3. Start the Logger Pro program on your computer. Open the file "09 Mole Ratio" from the Advanced Chemistry with Vernier folder. Obtain about 200 mL of each of the reactant solutions, NaOCl and Nas203. Measure out precisely 25.0 mL of the 0.50 M NaOCI solution. Pour this solution into a 4. 5. Styrofoam cup and nest the cup in a beaker to help stabilize the cup (see Figure 1). 6. Immerse the tip of the Temperature Probe in the Styrofoam cup of NaOCl solution. 7. Measure out precisely 25.0 mL of the 0.50 M Na2S2O3 solution. Note: Do not mix the two solutions yet 8. Click Collect to begin data collection. Let the program gather and graph a few initial temperature readings, and then add the Na S2O3 solution. Gently stir the reaction mixture with the Temperature Probe. 9. Data collection will stop after 3 minutes. You may click Stop to end data collection before three minutes have passed, if the temperature readings are no longer changing. 10. Examine the graph to calculate and record the maximum temperature change.
10. Examine the graph to calculate and record the maximum temperature change. a. To determine the highest temperature, click the Statistics button, . The minimum and maximum temperatures are listed in the statistics box on the graph. It may be necessary to examine the graph to determine the initial temperature, if the minimum temperature is not suitable b. Open Page 2 of this experiment file by clicking on the Next Page button, The table in this file is already set up for you to enter data. In the first line in the table, enter the volume of hypochlorite for the trial you just completed, as well as the temperature change in °C. c. Return to Page 1 of the experiment file. 11. Rinse out and dispose of the reaction mixture as directed. 12. Repeat the necessary steps to continue testing various ratios of the two solutions, keeping the total volume at 50.0 mL, until you have three measurements on either side of the ratio that produced the greatest temperature change 13. Print a copy of your final Page 2 graph (change in temperature vs. volume of hypochlorite). 9 -2 Advanced Chemistry with Vernier
Determining the Mole Ratios in a Chemical Reaction DATA TABLE Volume S203 Temperature change Volume OcIF (mL) (mL) DATA ANALYSIS 1. Determine the whole number mole ratio of the two reactants. Use the information in the raph you created on Page 2 of the experiment file (temperature change vs. volume of ypochlorite)
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Answer #1

As per the data provided,

Volume of OCl- Volume of S2O3- Moles of S2O3- (Molarity x Volume in liters) Moles of OCl- (Molarity x Volume in liters) Molar ratio S2O3-:OCl- Temperature change
20 mL or 0.02 L 30 mL or 0.03 L 0.5 x 0.03 = 0.015 moles 0.5 x 0.02 = 0.010 moles 0.015/0.010 = 3:2 6.5
30 mL or 0.03 L 20 mL or 0.02 L 0.5 x 0.02 = 0.010 moles 0.5 x 0.03 = 0.015 moles 0.010/0.015 = 2:3 10.1
10 mL or 0.01 L 40 mL or 0.04 L 0.5 x 0.04 = 0.020 moles 0.5 x 0.01 = 0.005 moles 0.020/0.005 = 4:1 6.3
40 mL or 0.04 L 10 mL or 0.01 L 0.5 x 0.01 = 0.005 moles 0.5 x 0.04 = 0.020 moles 0.005/0.020 = 1:4 2.6

5 mL or 0.005 mL

45 mL or 0.045 mL 0.5 x 0.045 = 0.0225moles 0.5 x 0.005 = 0.0025 moles 0.0225/0.0025 = 9:1

1.4

As the highest temperature change is brought about by the ratio of 2 mol of  S2O3- : 3 mol of OCl-

This the correct molar ratio

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