Limiting Reactant in a Mixture of Two Soluble Salts Lab Report

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. Introduction: In a chemical reaction, there are factors that affect the yield of products. Some being the amount of starting materials and the percent yield of the reaction. In this experiment we will be trying to determine the limiting reagent in the chemical reactions. We will be testing a known salt mixture and an unknown mixture. Well will be able to determine the limiting reagent by using a centrifuge. This allows us to separate the solid with the liquid in the reaction. The liquid sitting on top is called the supernatant. We will then test the two to see if any reactions occur. Based on the reaction we will be able to tell what the limiting reagent is in the reaction. Marterials: - 400mL beaker Stirring rod Hot plate 250mL beaker Graduated cylinder Procedure: 1. Prepare the salt mixture. a. Measure the mass of beaker 1 and record. Transfer 1g of salt mixture in beaker and measure the combined mass. b. Adjust pH of deionized water. Fill a 400mL beaker with deionized water. Test with pH paper. If water is acidic add drops of 6M NH3. c. Mix deionized water and salt. Add 150mL of the deionized water to the salt mixture in beaker 1. Stir with stirring rod for 2-3 minutes and allow precipitate to settle. 2. Digest the precipitate a. heat . cover the beaker and heat with hot plate. Do not go over 75 degrees C for 15 min. Periodically stir. b. Cool. after 15 min,remove heat and allow to cool. c. Wash water, heat 30 mL of deionized water for wash water. 3. Set up a gravity filtering apparatus. a. Place your initials and “trail 1” on a piece of wutnen NO. 42, fold and tear off its corner. Measure and record its mass. Seal filter paper into the filter funnel with a small amount of deionized water. 4. Withdraw and save supernatant. a. Once the precipitate has cleared in part 2b. Use a dropping pipet to withdraw enough supernatant to half fill two 75 mm test tubes labeled 1 and 2. Save for part b. 5. Filter the CaC2O4 x H2O precipitate. a. While the remaining solution of the salt mixture from part A4 is still warm, quantitatively transfer the precipitate to the filter. Transfer any precipitate on the wall of the beaker to the filter with the aid of a rubber policeman. Wash anything remaining . 6. Dry and measure the amount of CaC2O4 x H2O precipitate. Remove the filter paper and precipitate from the funnel. Air-dry the precipitate on the filter paper. Determine combined mass. 7. Formula of the precipitate. 1. - Clarify the supernatant. Centrifuge the two collected supernatant samples from part A4. Test for excess C2O4 Test for excess Ca2+ Schemes: Data: Data of known Mass 250 mL beaker 110.151g Salt mixture + beaker 111.198g Salt mixture + Beaker + dw 150 mL pH deionized water 224.816g 7 Salt mixture 1.45g mass filter paper 1.197g Beaker for precipitate 115.602g filter paper + precipitate 2.153g Precipitate 0.986g Data of unknown Mass 250 mL beaker 110.464g unknown salt mix 1.123g weighing paper 0.425g beaker + unknown 111.152g beaker +unknown + solution 245.725g Filter paper 1.178g Precipitate 0.148g Data analysis: Reactions: CaCl2 x 2 H2O + K2C2O4 x H2O ---> CaC2O4 x H2O + 2KCl + 2H2O Ionic: Ca2+ + 2Cl- + 2K+ + C2O42- + 3H2O ----> CaC2O4 x H2O + 2K+ + 2Cl- + 2H2O Net ionic: Ca2+ + C2O42- + H2O ----> CaC2O4 x H2O Known: test tube 1 CaCl2 → formed a precipitant (C2O42- excess + Ca+ is the limiting reactant) Unknown: test tube 1 CaCl2 → formed a precipitant Both had no reaction with K2C2O4 Calculations: Above is the calculations done with the known and unknown salt mixtures we tested. For the known, in the product CaC2O4 x H2O we found that Ca was the limiting reactant. Known Theoretical yield - 0.2985g CaC2O4 x H2O %yield = 0.2985/0.986 x 100 = 30.27% Supernatant test - CaCl2 - Formed precipitate Unknown Mass of mixture - 1.123g Mass of excess - 1.0146g Mass of limiting reagent - 0.1084g Mass of precipitate - 0.148g %mass= 1.014/1.123 x 100 = 90.3% Results and discussion Based on our data and our calculations we were able to execute this lab experiment with accuracy and precision. We were able to find the limiting reagent in each of the given compounds. For the known substance, which was CaC2O4 x H2O, we found that the limiting reagent was Ca. We found this by creating a balanced chemical equation and using their molar ratios to see what there was an excess of in the reaction. With the limiting reagent we were able to find the theoretical yield of the compound. The theoretical yield came out to be 0.2985g of CaC2O4 x H2O. With this number we are then able to find the percent yield which came out to be 30.27%. For the unknown we had to take a different approach to finding the limiting reagent. Shown above is the equation we used to help us find the %mass if the products. The mass of the mixture(1.123g) is equal to the mass of the excess(1.0146g) plus the mass of the limiting(0.1084g) is also equal to the mass of the precipitate(0.148g). Using this equation we were able the find the %mass which was 90.3%. Conclusion: In this experiment we were given a known and unknown salt mixture. We were then prompted to find the limiting reactants. We used a number of different methods to find these results. For example we used a centrifuge to separate the excess and limiting reactants. By doing this we were able to test the supernatants for see if a reaction occurred. When the reaction occurred we were able to tell what the limiting reactant was. The excess would not have had a reaction due to the fact that the substance was already used. This was interesting to see rather than just doing a math calculation in our books. Finally with the results we found, we used them to determine other sorts of data such as %yield and %mass.