Embedded Files
Limiting the Reaction
Limiting the Reaction
This page contains both the Virtual Experiment 9. We will be completing the Virtual experiment this semester.
The assignments that must be completed for this laboratory assignment are:
Complete the pre writing and the pre laboratory problems before coming to the class meeting.
Complete the laboratory report using the class data. Both sections must be complete with data and calculations. Complete all questions in the Laboratory Report.
Complete the post laboratory problems and conclusion paragraph
Discussion:
Discussion:
In a double replacement reaction, two reactants are necessary for the reaction to occur. When two reactants are present, the quantity of each can affect the extent of the reaction. When both reactants are in the exact ratio needed so that each reactant is completely consumed, then the amount of product is directly related to amount of either or both of the reactants. However, what often happens in reactions is that one reactant is in a smaller amount needed for the complete reaction of the second reactant or there is more of the second reactant than actually needed. This reactant is said to be in excess and some of this reactant will remain in the reaction vessel. However, the reactant is the smaller amount is said to be the limiting reactant because it will be completely consumed and thereby limits the amount of product that can be made. We use stoichiometry to determine the amounts of product being formed, but we can also determine the amounts of each reactant needed.
The calculations
The first reaction above is a traditional double replacement reaction which contains two ionic compounds which exchange partners; Another example of a double replacement reaction is: AgNO3 (aq) + NaCl (aq) = AgCl (s) + NaNO3 (aq) where the formation of the insoluble ionic compound of silver chloride drives this reaction toward completion. If an insoluble compound is formed as a product, the reaction will most likely occur. The coefficients in the reaction give the relative amounts of moles of each species. In this case you can not make any more silver chloride than number of moles of silver nitrate present.
With the balanced chemical reaction complete, the amounts of reactants and products can be determined using the molar mass of each compound in the reaction and molar ratios as determined by the coefficients of the balanced chemical reaction.
The number of grams is measured in the laboratory and converted to the number of moles using the molar mass of the compound.
massA divided by molar mass of A (MM) = moles of A
gA x 1mol/ MM in g = molA
The molar ratio from the balanced chemical reaction allows us to determine the theoritical amount of product created if the reaction ran perfectly.
moles of product = moles of reactant x (molar ratio)
where the molar ratio = coefficient of moles of product from balanced reaction divided by the coefficient of the moles of reactant from the balanced chemical reaction.
From the theoretical number of moles produced, the theoretical yield or mass of the product can be determined by using the molar mass of the substance produced.
moles of the product determined above x molar mass of the product = mass of product (theoretical yield)
The percent yield of a reaction is defined as the experimental mass (mass collected by the student) divided by the theoretical mass, times 100. The percent deviation is the same as the percent error. The percent yield + the percent deviation = 100 %
Reactants, Products and Leftovers
Chemical Reactions
Limiting Reactants
This simulation will help you balance simple reactions and determine the limiting reactant.
Pre laboratory Writing assignment
Pre laboratory Writing assignment
The student should complete a pre laboratory writing assignment that contains a purpose, a summary of the background or discussion and a summary paragraph of the procedure explaining how the assignment will be completed and what data must be collected. This assignment will only be accepted before the laboratory session begins. No late assignments will be accepted.
Pre laboratory assignment
Pre laboratory assignment
The pre laboratory problems are questions about the laboratory assignment, background, definitions or procedure and calculations. These problems should be completed BEFORE coming to the class where the assignment will be discussed.
Procedure and Report
Procedure and Report
The laboratory report of the assignment contains questions, data collection, calculation and reporting of results.
Students can use the data collected by the class to complete the laboratory assignment.
The reaction
The reaction
The hydrated salt is first dissolved in water, when this occurs the water that is part of the salt is removed and the ionic compound dissolves in water.
Na3PO4 . 12 H2O (solid) → Na3PO4 (aqueous) → 3 Na+ + PO43- in solution
SrCl2 . 6 H2O (solid) → SrCl2 (aqueous) → Sr2+ + 2 Cl- in solution
The unbalanced reaction would then be a double replacement reaction
Na3PO4 (aq) + SrCl2 (aq) → Sr3(PO4)2 (solid) + NaCl (aq)
Once this reaction is balanced the coefficients will allow us to determine the molar ratios and complete the process of stoichiometry.
Video of sodium carbonate with strontium chloride (not the same but very similar) https://youtu.be/pWZoX3tE0qw or https://youtu.be/e-QuDh6NiTg
Calculating limiting reactants:
Bozeman Science: https://youtu.be/LicEaaXhlEY
Tyler DeWitt: https://youtu.be/nZOVR8EMwRU and https://youtu.be/N0dTXcoHI-I
This video show the reaction of strontium chloride hexahydrate with sodium phosphate dodecahydrate and the measurement of the amounts used in the reaction for Beaker 1.
Exp 9 reaction.mp4
Beaker I - use 1.0 gram of sodium phosphate dodecahydrate and 1.5 g of strontium chloride dihydrate dissolved and digested to produce the precipitated product to determine the percent yield of the reaction from the theoretical yield of the product of the reaction.
If 1.0 grams of Na3PO4 is used, then we can determine the number of grams of SrCl2 needed to exactly react to produce only the two products of the reaction and using up both reactants completely with no excess reactants left over in the solution mixture:
1.000 grams Na3PO4 x (1 mol / 380.06 g) x ( 3 mol SrCl2 / 2 mol Na3PO4) x 194.54 g / 1 mol) = 0.77 g SrCl2
since 1.5 g is added, that means that almost 0.7 grams will not be reacted to produce the strontium phosphate product. If more sodium phosphate is added, we could make more strontium phosphate.
This should mean that the SrCl2 is the compound in excess and when the Na3PO4 reagent is added to the collected fluid from the filtration, we should see a precipitate only when the sodium phosphate is added.
The precipitate is collected, dried and weighed to determine the % yield of the reaction.
The theoretical moles of Sr3(PO4)2 can be calculated from the mass of the limiting reactant.
grams of limiting reactant divided by molar mass of compound = moles of limiting reactant
moles of limiting reactant x molar ratio of Sr3(PO4)2 to limiting reactant = theoretical moles of Sr3(PO4)2
theoretical moles of Sr3(PO4)2 x molar mass = theoretical yield of Sr3(PO4)2
the actual yield is the weight of the collected precipitate
the percent yield is the actual yield divided by the theoretical yield of Sr3(PO4)2 times 100
Exp 9 digestion and filtering.mp4This video shows the progression of the reaction during digestion and calculation of the theoretical moles of product for Beaker 1.
Exp 9 balancing reaction.mp4This video shows the balancing of the reaction of strontium chloride hexahydrate with sodium phosphate dodecahydrate
Exp 9 reaction and filtering.mp4This video shows the filtering of the precipitate of the reaction.
Exp 9 testing for limiting.mp4This video shows how we determine the limiting reactant using solutions of the reactants.
Exp 9 calculations for beaker 1.mp4Continuing calculations of theoretical yield for Beaker 1.
Beaker II will use a pre-mixed sample that contains both salts. The student will weigh between 1.5 - 2.0 grams of this mixture and perform the same procedure to determine the percent by mass of each salt in the mixture. The most difficult idea to determine is which component of the mixture is in a greater amount. In both parts of the experiment, we will check the collected fluid that passes through the filter paper for the presence of one of the two compounds used in the solution with the 1 M solutions of the two reactants.
Exp 9 beaker 2 reaction.mp4Discussion of Beaker 2 and the reaction.
Exp 9 Beaker 2.mp4Discussion of the calculations for Beaker 2
Post Laboratory Problems and Conclusion paragraph
Post Laboratory Problems and Conclusion paragraph
The post laboratory problems contains questions that determine your understanding of the laboratory assignment.
While the conclusion paragraph is a summary of what you learned in the assignment. The conclusion paragraph should not just be "this is what I learned" statement but a summary of the assignment, analysis of the data and results of the assignment and critical thinking about what was learned and why it is important. The RERUN method is described in Appendix D.
Report abuse