Finding the Ratio of Mole of Reactants in a Chemical Reaction

Title finding the Ratio of Mole of Reactants in a Chemical Reaction utilization of Lab To find the coefficients of twain chemical reactants that appears in a equilibrate chemical equation victimization the continuous variations method. Pre Lab Questions 1. 2AgNO3(aq) + K2CrO4(aq) 2KNO3(aq) + Ag2CrO4(s) 2. There is full to make a valid conclusion because, on the graph, one loafer clearly see that the deuce lines intersect. Thus, one can also checker the breakwater balance. The counterspy ratio is approximately 21 (AgNO? to K? Cr? ). Procedure 1.Obtain 175mL of the bleach solution in a average 400mL beaker and 175mL of dissolving agent B in another beaker. 2. Measure the temperature of both solutions and record. The solutions should be the same temperature. If they are not, you will need to make a castigation for the temperature difference. 3. exploitation a clean 10mL graduated cylinder, bank bill 5. 0m of NaClO and pour the solution into a Styrofoam loving cup. Usi ng a clean 50mL graduated cylinder, measure 45. 0mL of Solution B and add this to the Styrofoam cup. 4. Stir with a thermometer, and record the maximum temperature reached of the lowest solution in your data t equal. . Pour the solution out, rinse the cup and thermometer, and repeat steps 1-4 using a different ratio of the two substances, always keeping the stack at 50. 0mL. 6. Continue testing unlike ratios until you use up at least three measurements on each boldness of the one that gave the greatest temperature difference. 7. Plot your data on a graph, ? T ? vs. Reactant volume (mL). 8. Draw two best fit straight lines, and situate where they intersect. Be sure to include the points at the 050mL and 500mL rations. If at any points do not f every close to the lines, repeat these measurements.Find the stoichiometric jetty ration of reactants from the line of converging on the graph. Data initial Temp of both 23. 5? Na2SO3 (mL) NaClO(mL) Max. Temp. (? ) 45. 0mL 5. 0mL 28. 9? 5. 0mL 45. 0mL 27. 5? 25. 0mL 25. 0mL 42. 3? 40. 0mL 10. 0mL 33. 4? 10. 0mL 40. 0mL 30. 9? 30. 0mL 20. 0mL 43. 5? 20. 0mL 30. 0mL 38. 6? Conclusion In this lab we plunge the coefficients of two chemical reactants that appeared in a balanced chemical equation using the continuous variations method. The mole ratio turned out to be 11.One can be able to find the correct mole ratio after plotting the results on a graph that showed the trend of alter in temperature. Theory The theory of this experiment was continuous variation because it gives the stoichiometric ratios for the correct chemical equation if one does them correctly. Error Analysis instead of keeping just one initial temperature for all the tests, to make it more than than accurate, we could work measured the initial temperature of the solutions before each reaction. The exactness of the thermometer could have altered our results.Related essay Measuring Reaction Rate Using Volume of Gas Produced LabAlso, there was an error in measuring the reactants using graduated cylinders. This more or less likely happened because graduated cylinders are not the close accurate way to measure a liquid. Post Lab Questions 1. contrastive temperature changes occur when mixing different volumes of the two reactants. The experiment was made so the volume of solution and the total number of moles of reactants was a constant for all ofexperiments, since temperature change was directly proportional to the amount of reactants. Therefore, the optimum ratio is the ratio of the volume of reactants.It consumed the greatest amount of reactants, formed the greatest amount of products, and generated the most heat and maximum temperature change 2. A constant volume of reactants moldiness be kept in all trials so that the subsequent changes in temperature can be compared. They are proportional to the amount of reactants consumed in the reaction. 3. Yes. The concentrations ofthe two solutions must be the same in order for the stoichiometric mole ratio of the reactants to be accurate after the results of the trial are graphed because heroconcentration is moles of solute divided by liter of solution. . A limiting reagent is reactants in a chemical reaction that will be completely bygone before all other reactants in the reaction are utilize up. This will also cause the reaction to stop. 5. Both measurements limit the precision of the data. The measurement of temperature limits the precision of the data to 3 significant figures. The thermometer utilize was able to be read to the tenths of a degree. On the other hand, the measurement ofvolume limits the precision of the data to 2 significant figures.The graduated cylinders used, only could measure volume to the nearest mL and anything after that was an estimate. 6. The limiting reagent on the upward(a) sloping line of the graph is Na? SO?. The limiting reagent along the downward slope line is NaClO. 7. Other physical properties that could be us ed in the method of continuous variation, besides temperature change, could be color change, the surge of a precipitate that forms, and/or the volume of a gas produced. 8.It is more accurate to use the point of intersection of the two lines to find the mole ratio rather than the ratio associated with the greater temperature change because one most likely did not use the exact mole ratio. Thus, the point of intersection is a more likely an average. Also, the maximum amount of each reactant force not have been used to get the highest temperature the highest temp wouldnt necessarily have the highest mole to more ratios. 9. The average of the temperatures would be used as the initial temperature.Graph Calculations The intersection point was at approximately 28. 6 mL Na? SO3? and 21. 4 mL NaClO. Moles Na? SO3? = 0. 5 M(. 0286 L) = . 0143 moles Na? SO3? Moles NaClO = 0. 5 M(. 0214 L) = . 0107 moles NaClO .0143 moles Na? SO3? /. 0107 = 1. 34 moles Na? SO3? which rounds to the nearest who le number, 1 . 0107 moles NaClO/. 0107 = 1 mole NaClO The stoichiometric mole ratio is 11 for the reactants. The actual stoichiometric mole ratio for the reaction is 11.