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Experiment 8 Prelaboratory Assignment Limiting Reactant Desk No. Date Lob Sec. Nome 1. The limiting reactant is determined in this experiment. a. What are the reactants (and their molar masses) in the experiment? b. How is the limiting reactant determined in the experiment? 2. Experimental Procedure, Part A.2. What is the procedure and purpose of "digesting the precipitate"? 3. Two special steps in the Experimental Procedure are incorporated to reduce the loss of the calcium oxalate precipitate. Identify the steps in the procedure and the reason for each step. Experiment 8 129
4. a. A sample of a CaC1, 2H,O/K,C,OH,O solid salt mixture is dissolved in -150 mL of deionized water previously adjusted to a pH that is basic. The precipitate, after having been filtered, was air-dried and weighed. Data for Trial were obtained as shown. Complete the following table. (See Remart Sheet) Record calculated values with the cor- rect number of significant figures. Calculation Zone A. Precipitation of CaC,OH,O from the Salt Mixture 0.879 Data Analysis, 1. 3. Mass of salt mixture (g) 4. Mass of filter paper (8) 5. Mass of filter paper and CaC,0,*H,0 (g) 6. Mass of air-dried CaC,0 •H,0 (8) 2.180 Data Analysis, 3. B. Determination of Limiting Reactant 1. Limiting reactant in salt mixture 2. Excess reactant in salt mixture CaCl 2H.0 Data Analysis, 4. Data Analysis 1. Moles of CaC,0,H,O precipitated (mol) Show calculation 2. Moles of limiting reactant in salt mixture (mol) See equation 8.1. 3. Mass of limiting reactant in salt mixture (8) Show calculation 4. Mass of excess reactant in salt mixture (R) Equals mass of salt mixture minus mans of limiting reactant. Show calculation 5. Percent limiting reactant in salt mixture (96) Show ealculation 6. Percent excess reactant in salt mixture (56) Data Analysis, 5. 4. b. For Trials 2 and 3, the percent CaCl 2H 0 in the mixture was 39.7% and 28.4% respectively. a. What is the average percent of CaCl-2H,O in the sample? Data Analysis, B. b. What are the standard deviation and the relative standard deviation (RSD) for the percent of CaCl, 2H,O in the sample? Data Analysis, C and D. 130 Limiting Reactant
According to equation 8.1, the moles of CaCl 2H 0 that react equals the moles of K.C.O-H,0 (the limiting reactant) that react equals the moles of Cac,OH,O that precipitate. Therefore, the mass of CaCl 2H,0 (the excess reactant) that reacts is mass CaCl-2H,0 -0.194 g CacOH;0 Imol CaC,0,HO 146.12 g CaC,0,H, 1 mol CaC1,-2H,0 147.02 g CaCl, 2H,0 1 mol CaCO, H, 1 mol CaCl2H40 -0.195 g CaCl, 2H,0 reacted Since 0.293 g CaCl 2H,0 were calculated present in the original salt mixture, then the mass of excess CaC1, 2H,0 is (0.293 8 -0.195 g =) 0.098 g xs CaCl, 2H,0. Procedure Overview: In Part A, a measured mass of a solid CaCl2H,O/K.C,OH,O EXPERIMENTAL salt mixture of unknown percent composition is added to water. The precipitate that PROCEDURE forms is digested, filtered, and dried, and its mass is measured. Observations from tests on the supernatant solution in Part B determine which salt in the mixture is the limit. Supematant, the clear solution that ing reactant. An analysis of the data provides the determination of the percent compo- exi ofer the precipitate has settled sition of the salt mixture. Two trials are recommended for this experiment. To hasten the analyses, measure the mass of duplicate unknown solid salt mixtures in clean 150 or 250-ml, beakers and simultaneously follow the procedure for each. Label the heakers accordingly for Trial I and Trial 2 to avoid the intermixing of samples and solutions. Obtain about 2-3 g of an unknown CaCl, 2H,O/K.C.O.H.O salt mixture. Be aware of the number of significant figures when recording data. A. Precipitation of CaC,0,H,O from the Salt Mixture Data Analysis, A 1. Prepare the salt mixture. a. Mass of salt mixture. Measure the mass (0.001 g) of heaker I and record on the Report Sheet for Trial 1. Transfer - g of the salt mixture to the beaker, measure, and record the combined mass. Repeat for Trial 2, beaker 2. b. Adjust pH of deionised water. Fill a 400-ml beaker with deionized water. Test with pH paper. If the water is acidic, adjust it to be just basic with drops of 6 M NH,. If already basic to pH paper, then no addition of NH, is necessary. c. Mir deionised water and salt. Add-150 ml of the deionized water from Part A.lb to the salt mixture in beaker 1. Stir the mixture with a stirring rod for 2-3 minutes and then allow the precipitate to settle. Leave the stirring rod in the beaker. Repeat for Trial 2 2. Digest the precipitate." a. Heat Cover the beaker with a watch glass and warm the solution on a hot plate (Figure 8.1. page 126) to a temperature not to exceed 75°C for -15 minutes. Periodically stir the solution and, in the meantime, proceed to Part A.3. b. Cool. After -15 minutes, remove the heat and allow the precipitate to settle; the solution does not need to cool to room temperature. Calcium oxalate does not precipitate in an acidic solution because of the formation of HC, on ion that does not precipitate with Cod Digesting a precipitate in warm water promotes the dissolving of the smaller precipitate particles fol lowed by their reprecipitating on the existing larger porticles. The result is a more efficient filering process Experiment 8 125
Spectator lor corions or anions that do not participare in any observable of detectable chemical reaction Nel lonic equation on equation that includes only those ions that participate in the observed chemical reaction, aho presented in Experiment Presenting only the ions that show evidence of a chemical reaction (.e., the forma in a recinitate arby removing the spectator ions (ie., no change of ionic for during the reaction), we have the net ionic equation for the observed reaction: Calaq) + CO Tag) + H.O(1) CaC,OH,O(s) (8.3) Calcium oxalate monohydrate is thermally stable below -90'C but forms the anhydrous salt, Caco at temperatures above 110°C. Therefore, one mole of Cat (from one mole of CaC12H,0 molar mass = 147.02 g/mol) reacts with one mole of C.0,- (from one mole of K,C,OH,O, molar mass = 184.24 g/mol) to produce one mole of Cac,OH,O (molar mass = 146.12 g/mol). If the calcium oxalate is heated to temperatures greater than 110°C for drying, then anhy- drous Cac.o. (molar mass = 128.10 g/mol) is the product. In Part A of this experiment the solid reactant salts CaCl2H.O and K.C.0/H,0 form a heterogeneous mixture of unknown composition. The mass of the solid mixture is measured and then added to water-insoluble CaC,OH,O forms. The Cac,OH,O precipitate is collected via gravity filtration and dried, and its mass is measured. The percent composition of the salt mixture is determined by first testing for the limiting reactant. In Part B. the limiting reactant for the formation of solid cal- cium oxalate monohydrate is determined from two precipitation tests of the final reaction mixture from Part A: (1) the mixture is tested for an excess of calcium ion with an oxalate reagent-observed formation of a precipitate indicates the presence of an excess of calcium ion (and a limited amount of oxalate ion) in the salt mix- ture; (2) the mixture is also tested for an excess of oxalate ion with a calcium reagent-observed formation of a precipitate indicates the presence of an excess of oxalate ion (and a limited amount of calcium ion) in the salt mixture. Calculations The calculations for the analysis of the salt mixture require some attention. "How do I proceed to determine the percent composition of a salt mixture of CaCl, 2H,0 and K.C.O-H,O by measuring only the mass of the CaC,OH,O precipitate? Example: A 0.538-g sample of the salt mixture is added to water and after drying (to less than 90°C) 0.194 g of Cac,OH,O is measured. Tests reveal that K.C.O-H,O is the limiting reactant. What is the percent composition of the salt mixture? Since K.C.OH,is the limiting reactant, how many grams of the excess CaC1,2H,0 were in the salt mixture? Solution: Since K.C.O,H,O is the limiting reactant, then, according to equation 8.1, the moles of K.C.O-H,0 in the salt mixture equals the moles of CaC,0,H.0 formed. Therefore, the calculated mass of K.C.O-H,O in the original salt mixture is 6. I wol CaC20, HẠO grams K.C.O,H,0 -0.194 g CaC,0,H,0 x 146.12 g CaC,,H,O 1 mol KC,OH,O 184.24 K.C.O.-H,0 Imol CaC,0,H, 0 1 mol KC,0,H,0 0.245 g K.C.O-H,O in the salt mixture. The percent by mass of K.C.O-H,0 in the original salt mixture is -0.245 g K.C.O.H.O -X 100 = 45.5% K.C.0H,0 Data Analysis, A SK.C_0H,0 = 0.538 g sample The mass of the CaCl, 2H,0 in the salt mixture is the difference between the mass of the sample and the mass of K.C,0H,O or (0.538 g - 0.245 g ) 0.293 g. The percent by mass of CaCl-2H 0 in the original salt mixture is 0.5388 -0.2458 x 100 = 54.5% CaCl, 2H,0 CaCl-2H0 = 0.538 g sample 124 Limiting Reactant
Experiment 8 Limiting Reactant Scimol/Science Source Calcium oxalate crystals contribute to the formation of kidney stones. OBJECTIVES • To determine the limiting reactant in a mixture of two soluble salts • To determine the percent composition of each substance in a salt mixture The following techniques are used in the Experimental Procedure: TECHNIQUES Percent composition: the mos ratio of component of a midwe or compound to the total mass of the sample times 100 octual yield) x 100 Two factors affect the yield of products in a chemical reaction: (1) the amounts INTRODUCTION (moles) of starting materials (reactants) and (2) the percent yield of the reaction. Percantil Many experimental conditions, for example, temperature and pressure, can be adjusted to increase the yield of a desired product in a chemical reaction, but because chemicals theoretical yield/ react according to fixed mole ratios (stoichiometrically), only a limited amount of product can form from measured amounts of starting materials. The reactant determin- ing the amount of product generated in a chemical reaction is called the limiting reac- Stoichiometricely: by a study of a tant in the chemical system. chemical reaction using a balanced To better understand the concept of the limiting reactant, let us look at the reac equation tion under investigation in this experiment, the reaction of calcium chloride dihy- drate, CaCl, 2H,0, and potassium oxalate monohydrate, K,C,0,H,O, in an aqueous solution. CaCl-2H,O(aq) + K.C.OH,O(aq)- Cac.0.H.0(s) + 2 KCl(aq) + 2 H.(I) (8.1) Calcium oxalate monohydrate, Caco H., the product, is an insoluble com- pound, but is found naturally in a number of diverse locations. It is found in plants, such as rhubarb leaves, agave, and (in small amounts) spinach, and is the cause of most kidney stones. In small doses, it causes a severe reaction to the lining of the digestive tract. However, the handling of calcium oxalate in the laboratory is safe, so long as it is not transferred to the mouth. For the reaction system in this experiment, both the calcium chloride and potas- sium oxalate are soluble salts, but the calcium oxalate is insoluble. The lonie equation lonic equation: A chemical equation for the reaction is that presents lonic compounds in the form in which they exist in aqueous Cal(aq) + 2 CH(aq) + 2 K*(aq) + C,0, (aq) + 3 H,O(1) - solution. See Experiment 6. C.C.O.-H,0(s) + 2 Cl(aq) + 2 K*(aq) + 2 H2O(l) (8.2) Experiment 8 123
According to the chegg guidelines I have solved first four parts/ subparts of the assignment due to the paucity of time i.e 2 hours only to answer the questions -