Using experimental measurements to determine the stoichiometric ratio in the equation, the optimum ratio.
Data Tables:
Combinations
Initial Temperature
0 mL NaClO 50 mL Na2SO3
10 mL NaClO 40 mL Na2SO3
22.05 °C
22.05 °C
20 mL NaClO 30 mL Na2SO3
30 mL NaClO 20 mL Na2SO3
22.05 °C
22.05 °C
40 mL NaClO 10 mL Na2SO3
50 mL NaClO 0 mL Na2SO3
22.05 °C
22.05 °C
Observed temperature after reaction
Change in temperature
22.0 °C
28.3 °C
.05 °C
6.25 °C
34.0 °C
36.9 °C
11.95 °C
14.85 °C
28.5 °C
22.1 °C
6.45 °C
.05 °C
Graphs:
Last Reactant volume ratio on the graph is suppose to be 50 mL NaClO and 0 mL Na2SO3
Conclusion: The optimum ratio of NaClO and Na2SO3 should have a change of 17.5 degrees Celsius. The reactant volume ratios is close to 30 ml of NaClO and 20 mL of Na2SO3. According to the graph it is 31 mL of NaClO and 19 mL of Na2SO3. This is the ratio that will produce the greatest change in temperature, consume the greatest amount of reactants, and form the greatest amount of product. The mole to mole ratio is 3 mol NaClO to 2 mol Na2SO3. Every 3 mol of NaClO for 2 mol Na2SO3 to produce the products. This was calculated by setting the .5M NaClO and .5M Na2SO3 to there volumes (ml to liters) and solve for the moles. .0155 moles of NaClO and .0095 moles of Na2SO3 was calculated.
Discussion of Theory: Reacting the different volume ratio of Na2SO3 and NaClO gave off different amounts of heat. The temperature was directly proportional to the amount of reactant consumed during the reaction. The numbers could be compared because the volume and the total number of moles were kept constant throughout the experiment. After the change in temperatures and the different volume ratios were graphed the optimum ratio could be found. Two best fit lines were drawn and one was sloping upwards and the other one was sloping downwards. These lines helped determine the optimum ratio by estimating an experimental value we could have got based on the data. An experiment was not conducted at every possible volume ratio, so the graph helps to estimate a possible ratio that will produce the greatest amount of products, generate the most heat (maximum temperature change), and consume the greatest amount of reactants.
Analysis Questions:
1. Why did you have to keep a constant volunte of reactants? You have to keep a constant volume of reactants in order to compare the temperatures. The temperatures indicate how much of the reactant is consumed. 2.What is meant by the term "limiting reagent"? The term limiting reagent refers to the substance that is totally consumed at the end of a chemical reaction. The amount of products produced is limited because the reaction cannot take place without this reagent. 3.Which measurement limits the precision of your data: temperature or volume? Explain. The measurement of the temperature limits the precision because the thermometer measures by 1 degrees Celsius, while the 10 Luated cylinders measures by .1 mL. Since mLe measurement of the temperature is by larger increments the data may seem to vary widely when the numbers are actually close together. 4.Which reactant is the limiting reagent along the upward sloping line of your graph? Which is the limiting reagent along the downward sloping line? NaClO is the limiting reagent along the upward sloping line of the graph because as the mL of NaClO increase from 0 mL to 20 mL the amount of heat produced increases. Na2SO3 is the limiting reagent along the downward sloping line because as the mL of Na2SO3 decrease 20 to 0 mL the amount of heat produced decreases as well. 5.What physical properties, other than temperature change, could use the method of continuous variations? Color and the amount of precipitate formed could use the method of continuous variation. 6.Why is it more accurate to use the point of intersection of the two lines to find the mole ratio, rather than the ratio associated with the greatest temperature change? It is more accurate to use the point of intersection of the two lines because an experiment was not conducted for every possible ratio, so a experimental value was not obtained for the optimum ratio during the experiment. The optimum ratio can not be determined with the ratio associated with he greatest temperature change because slight changes can affect the temperature. Therefore, if you use two lines it can find the optimum ratio taking into account for both limiting reagents (on both slopes).