Lab 5: Limiting Reagent Study Lab Report

Lab 5 –Limiting Reagent Study– Lab Report CHM 2045L -Dr. Yavuz Petrowski The statement is: “I certify that this is my own work and I understand that if I am found to be in violation of the honor code, I will be subject to the highest penalty” Name: Raul Garcia Znumber: Z23809105 PRELAB Background: Please complete the following sections. Purpose of the experiment: In your own words, the purpose of the experiment including a brief description of the lab and the reason for doing the experiment (what we are trying to determine and how we will go about determining it). Explain the concepts we are investigating; explain why and how they are useful, and how those concepts apply to the experiment; show relevant equations if you need to. The purpose of this experiment is to learn more about limiting reagents and their importance/uses in chemical reactions. In the lab, one can determine the limiting reagent by combining solutions of cobalt chloride and sodium carbonate to create solid, cobalt carbonate. By doing this, one can forecast the limiting reagent by comparing the heights of the precipitating particles and using observations and liquid supernatant tests. Flow Chart: Complete the experimental procedure for this experiment by using the information provided in the powerpoint and 360 experimental videos in a flow diagram format. A flow chart indicates the procedure of how the experiment is performed. The flow chart must be written in the correct flow chart format (box and arrow). Handwritten flow charts are not accepted. See the example of the flow chart in the canvas module. Waste Collection: Type out the chemicals used in this experiment, their hazards if applicable, and any other safety concerns associated with this experiment. You can search for the Materials Safety Data Sheet document on the internet to list hazards and safety concerns. You can add an extra row if you need extra space. Chemicals Safety Hazard Cobalt chloride Can cause skin/eye irritation. Sodium phosphate Can cause skin/eye irritation, as well as throat irritation and cough. Barium chloride Can cause skin/eye irritation, as well as respiratory tract irritation. Data Process YOU MUST SHOW ALL YOUR CALCULATIONS TO RECEIVE CREDIT FOR THEM! Proper use of significant figures and units is part of the grading. PART I: for CoCl2 TEST-TUBE TEST-TUBE TEST-TUBE TEST-TUBE TEST-TUBE TEST-TUBE #1 #2 #3 #4 #5 #6 Concentration The volume 0.10 M 2.00 3.00 4.00 6.00 7.50 10.00 0.0002 0.0003 0.0004 0.0006 0.00075 0.00010 delivered (mL) Moles of CoCl2 For Na2CO3 Test-tube #1 Test-tube #2 Test-tube #3 Test-tube #4 Test-tube #5 Test-tube #6 for CoCl2 TEST-TUBE TEST-TUBE TEST-TUBE TEST-TUBE TEST-TUBE TEST-TUBE #1 #2 #3 #4 #5 #6 Concentration The volume 0.10 M 13.00 12.00 11.00 9.00 7.50 5.00 0.0013 0.0012 0.0011 0.0009 0.00075 0.0005 delivered (mL) Moles of Na2CO3 Volume Measurements Height of Test-tube #1 Test-tube #2 Test-tube #3 Test-tube #4 Test-tube #5 Test-tube #6 5.5cm 6.4cm 7.3cm 8.3cm 8.0cm 7.5cm Clear Clear Clear Clear Clear Clear Blue Blue-purpl Dark Purple Light Pink-purpl e purple Purple e precipitate formed (cm) Color of supernatant liquid observed. Color of precipitate observed. Calculations: Complete mol calculations for CoCl2 and Na2CO3 by using the information table given in the instructions file, and/or PowerPoint and 360 Lab video. Handwritten calculations are not acceptable. If the student submits a handwritten calculation, the assignment will be graded with a 4-point deduction. CoCl2 Calculations Test-Tube #1 Volume=2.00mL=0.002L Moles=0.1M×0.002L Moles of CoCl2=0.000200 mol Test-Tube #2 Volume=3.00mL=0.003L Moles=0.1M×0.003L Moles of CoCl2=0.000300 mol Test-Tube #3 Volume=4.00mL=0.004L Moles=0.1M×0.004L Moles of CoCl2=0.000400 mol Test-Tube #4 Volume=6.00mL=0.006L Moles=0.1M×0.006L Moles of CoCl2=0.000600 mol Test-Tube #5 Volume=7.50mL=0.0075L Moles=0.1M×0.0075L Moles of CoCl2=0.000750 mol Test-Tube #6 Volume=10.00mL=0.010L Moles=0.1M×0.010L Moles of CoCl2=0.00010 Na2CO3 Calculations Test-Tube #1 Volume=13.00mL=0.013L Moles=0.1M×0.013L Moles of Na2CO3=0.0013 mol Test-Tube #2 Volume=12.00mL=0.012L Moles=0.1M×0.012L Moles of Na2CO3=0.0012 mol Test-Tube #3 Volume=11.00mL=0.011L Moles=0.1M×0.011L Moles of Na2CO3=0.0011 mol Test-Tube #4 Volume=9.00mL=0.009L Moles=0.1M×0.009L Moles of Na2CO3=0.0009 mol Test-Tube #5 Volume=7.50mL=0.0075L Moles=0.1M×0.0075L Moles of Na2CO3=0.000750 mol Test-Tube #6 Volume=5.00mL=0.005L Moles=0.1M×0.005L Moles of Na2CO3=0.0005 mol Predictions of Limiting Reagent: This section can be completed while waiting for the precipitate to settle. Excess Co2+ or TEST-TUBE #1 CO3^2- TEST-TUBE #2 CO3^2- TEST-TUBE #3 CO3^2- TEST-TUBE #4 CO3^2- TEST-TUBE #5 CO3^2- TEST-TUBE #6 CO3^2- 0.0 0.0 0.0 0.0 0.0 0.0005 CO32- present? Moles of Co2+ in excess Moles of CO32- in TEST-TUBE #1 0.0011 TEST-TUBE #2 0.0009 TEST-TUBE #3 0.0007 TEST-TUBE #4 0.0003 TEST-TUBE #5 TEST-TUBE #6 0 0 2x10^-4 3x10^-3 4x10^-4 6x10^-8 7.5x10^-4 5x10^-4 excess Theoretical Yield of CoCO3 (moles) PART II: Verification of Limiting Reagent Observations of precipitate Test-tube #1 Test-tube #2 Test-tube #3 Test-tube #4 Test-tube #5 Test-tube #6 CLEAR CLEAR CLEAR CLEAR CLEAR Pink WHITE WHITE WHITE WHITE Clear Pink formation Reaction of supernatant liq. with Na3PO4 Reaction of supernatant liquid with BaCl2 Calculations: Complete mol calculations for excess CoCl2 and Na2CO3 by using the information table given in the instructions file, and/or powerpoint and 360 Lab video and include theoretical yield calculations. Handwritten calculations are not acceptable. If the student submits a handwritten calculation, the assignment will be graded with a 4-point deduction. Test tube #5 7.5x10^(-4) mol of CoCl2-7.5x10^(-4) Na2CO3=0 Test tube #6 1.0x10^(-3) mol of CoCl2-5.0x10^(-4) Na2CO3=5.0x10^(-4) mol of Co^(2+) in excess Test tube #1 1.3x10^(-3) mol of Na2CO3-2.0 x10^(-4) CoCl2 = 1.1 x10 ^(-4) mol of Co3^(-2) in excess Test tube #2 1.2x10^(-3) mol of Na2CO3-3.0 x10^(-4) CoCl2 = 9.0 x10 ^(-4) mol of Co3^(-2) in excess Test tube #3 1.1x10^(-3) mol of Na2CO3-4.0 x10^(-4) CoCl2 = 7.0 x10 ^(-4) mol of Co3^(-2) in excess Test tube 4 9.0x10^(-4) mol of Na2CO3-6.0 x10^(-4) CoCl2 = 3.9 x10 ^(-4) mol of Co3^(-2) in excess Test tube 5 7.5x10^(-4) mol of Na2CO3-7.5 x10^(-4) CoCl2 =0 Test tube 6 5.0x10^(-4) mol of Na2CO3-1.0 x10^(-3) CoCl2 = -5.0 x10 ^(-3) mol of Co^(2+) in excess Test tube 1 2.0x10^(-4) mol of CoCl2/1 x 1mol of CoCO3/1 mol of CoCl2 = 2.0x10^(-4) mol of CoCO3 Test tube 2 3.0x10^(-4) mol of CoCl2/1 x 1mol of CoCO3/1 mol of CoCl2 = 3.0x10^(-4) mol of CoCO3 Test tube 3 4.0x10^(-4) mol of CoCl2/1 x 1mol of CoCO3/1 mol of CoCl2 = 4.0x10^(-4) mol of CoCO3 Test tube 4 6.0x10^(-4) mol of CoCl2/1 x 1mol of CoCO3/1 mol of CoCl2 = 6.0x10^(-4) mol of CoCO3 Test tube 5 7.5x10^(-4) mol of CoCl2/1 x 1mol of CoCO3/1 mol of CoCl2 = 7.5x10^(-4) mol of CoCO3 Test tube 6 5.0x10^(-4) mol of CoCl2/1 x 1mol of CoCO3/1 mol of CoCl2 = 5.0x10^(-4) mol of CoCO3 CONSIDERATION FOR DISCUSSION & CONCLUSION As you write your discussion and conclusion paragraphs for the discussion assay, here are some tips and questions you should answer. These are in addition to the "Guidelines for a good discussion and conclusion" document in the introduction module. ●​ What is limiting reagent and how important in practical applications? Were your predictions for which ion was in excess verified by Part II of experiment? Explain. ●​ How did you determine the point at which a reagent from part I becomes the limiting reagent for the reaction? Does the molar ratio have an effect on which reagent is the limiting reactant? Explain. ●​ What is the difference between theoretical and percent yield? Why did you calculate the theoretical yield for your experiment? Would it be possible to calculate the experimental yield for your experiment and what information you would need? DISCUSSION & CONCLUSION ASSAY To start things off, let’s talk about limiting reagents. In a reaction, the limiting reagent is the reactant that determines the highest amount of product that may be formed. The moment the limiting reagent has been exhausted, the reaction will stop, regardless of how much of the other reactants exist. It is a fundamental concept in everyday uses such as in industrial manufacture, where reducing cost and wastage can be optimized by using all the reagents to their full capacity. In the lab, knowing the limiting reagent is applied to forecast the yield and efficiency of the reactions. In Part II of the experiment, we were able to verify the prediction of which ion was in excess. Through stoichiometric calculations, we established the limiting reagent, and what the results showed was that the ion we predicted to be in excess remained in excess following the reaction. To determine when a reagent will be the limiting reagent in the reaction, the amount of moles of each reactant was equated according to the molar ratio of the balanced chemical reaction. The molar ratio gives the proportion in which reactants are mixed. If a reagent is present in a smaller quantity than the ratio requires, then it will be the limiting reagent. In our example, by calculating the moles of each reactant and dividing them by the stoichiometric ratio, we were able to find the limiting reagent. The molar ratio is significant because it directly determines at what point one reagent is used up, preventing the reaction and determining the amount of product produced. In addition, the other important concepts are those of theoretical yield and percent yield. Theoretical yield refers to the amount of product which would be obtained if the reaction was 100% efficient from the limiting reagent and balanced equation. It is a theoretical value with no losses or side reactions. Percent yield, on the other hand, is a ratio of experimental yield to the theoretical yield in percentage value. This calculation assists in determining the efficiency of the reaction as well as any potential causes of losses. During this experiment, the theoretical yield was determined from the amount of limiting reagent and the molar ratio to estimate how much product would be produced. Determining the theoretical yield is important as it provides a standard against which the actual result of the experiment can be compared. To calculate the experimental yield, the amount of product that is actually made would need to be found out, most often by means of weighing or measuring concentration. Side reactions, incomplete reactions, or losses occurring during product recovery affect the experimental yield and are factors of interest to be considered when analyzing results. Overall, limiting reagent is key to predicting the outcome of chemical reactions. Molar ratio enables us to calculate the limiting reactant, which is key to the maximum product yield. Theoretical yield offers a baseline of comparison, while percent yield reveals how efficient an experiment was. Determining theoretical and experimental yields enables one to ascertain the effectiveness of a reaction and enhance it for subsequent experiments.