a) Title: The Establishment of a Chemical Equation Using the Method of Continuous Variation.
b) Purpose: the continuous variation method is used to establish a chemical equation by displaying on a line graph the change/difference in temperature caused by the composition of acids and bases in the solution.
II. Apparatus and Materials:
E)Pipet (10ml and 100 ml)
G)Plastic foam cup
III. Outline of Procedure: For each system, measure the quantities of the reactants with a buret. Measure the reactant having the smaller volume into a plastic- foam cup; measure the other into a smaller beaker. Measure the temperature of each solution and record the average of the two temperatures to the nearest 0.01˚C. Pour the solution contained in the beaker into the foam cup and stir with the thermometer. Continue stirring for one or two minutes, then record the maximum temperature obtained to the nearest 0.01˚C. Rinse the foam cup and beaker. Wipe, dry, and make another run. When all the runs are finished, plot a graph showing delta t versus volume of reactants. From the molarity of each reactant and the volume used, find the number of moles of each used to produce the maximum temperature.
nMOH+ H(11)X = .
A) Find the number of moles of base used per mole of acid.
B) find the heat of reaction (Delta H) in kilocalories (kcal) per mole of bas used, assuming the specific heat of the solution to be 1.00 Cal/degree-gram and the density of the solution to be 1.00 g/ml. use the delta t for the run nearest to that producing a maximum temperature in which there is an excess of acid used, or the value at the maximum of your curve.
C) Write a completely balanced equation for your reaction, including delta H.
IV. Concentration of based used 40 g/liter
Concentration of based used one mole/liter
Concentration of acid used one mole/liter
MI of base used| 0| 5| 10| 15| 20| 25| 30| 35| 40| 45| 50| MI of acid used| 50| 45| 40| 35| 30| 25| 20| 15| 10| 5| 0| Average initial temp.| 24.5| 24.3| 24.2| 24.1| 24.0| 23.9| 23.8| 23.7| 23.6| 23.5| 23.3| Max. temp.| 24.5| 25.4| 26.4| 27.8| 24.0| 30.4| 32.0| 32.3| 30.3| 27.1| 23.3| Δ temp.| 0| 1.1| 2.2| 3.7| 5.0| 6.5| 8.5| 8.6| 6.7| 3.6| 0|
V. Number of moles of base used to produce maximum temperature: .034 moles
Number of moles of acid used to produce maximum temperature: .016 moles
Number of moles of acid used per mole of base: 1:2.125
To get number of moles 34/1000 and 16/1000 ratio: 1:2, acid: base.
34 base =.034 M, 16 acid= .016 M
.016/.016= 1, .034/.016= 2.125
Acid: Base ratio
Percent Error: In order to get the percent error of the lab report you would first have to figure out the ratio. To get this you would have to divide the mole of acid and base by the lowest one. In this case, the lowest one is acid, which is .016. When you divide this by itself you get 1. When you divide .016 by .034 you get approximately 2.125, which turns out to be your ratio 1:2.125. To figure out you percent error you would need to take you actual-measured / actual x 100. The actual was 2, the measured was 2.125. Now you take 2-2.125= .125/2= .0625x100= 6.25. So your percent error comes out to be approximately 6.25%.
* Also you would need to figure out the heat of reaction (delta H). To figure this out you would need to take the Kilocalorie’s divided by the mole of base, which is .034. To figure out the Kcal you would need to take the highest point of the change in temperature which is 8.6 and multiply it by 50 and by 1 ( 8.6x50x1=430) then divide this by 1000 (430/1000= .430). .430 is your Kcal- so you...