Experiment 4 - Limiting Reactant PDF

Experiment - Four Limiting Reactant Objectives:  To determine the limiting reactant in a salt mixture.  To determine the percent composition of each substance in a salt mixture. Introduction: Two factors affect the yield of...

Experiment - Four Limiting Reactant Objectives:  To determine the limiting reactant in a salt mixture.  To determine the percent composition of each substance in a salt mixture. Introduction: Two factors affect the yield of product in a chemical reaction:  The amount of the starting materials (reactants): Chemicals react according to fixed mole ratio (stoichiomety), i.e., in exactly the correct amounts in the balance equation so that all reactants are used up during the reaction. When reactants are not mixed in the correct mole ratio (non-stoichiometric amounts), one reactant will be used up before the amount of the others are depleted. The reactant that is used up first limit the amount of product formed is called the limiting reactant Limiting reactant: it is the reactant that is consumed first and thus determines the amount of product formed.  The percent yield of the reaction: The amount of the product formed when the limiting reagent is completely consumed is called the theoretical yield of the product. Theoretical yield: is the maximum amount of the product that can be produced from the quantity of the limiting reactant used. Actually, the amount of product predicted by the theoretical yield is seldom obtains because of the side reaction and other complications. The actual yield of product is the amounts of the product that can be obtain experimentally. The actual yield is often less than the theoretical yield. The percent yield is represented as: Percent yield = actual yield x 100% Theoretical yield 45  To understand the concept of the limiting reactant, let us look at the reaction under investigation in this experiment, the reaction of Na3PO4.12H2O (molar mass=380 g/mol) and BaCl2.2H2O (molar mass = 244 g/mol) in aqueous solution. The molecular equation for this reaction is: 2 Na3PO4.12H2O + 3 BaCl2.2H2O → Ba3(PO4)2 + 6 NaCl + 30 H2O As the two reactant salts and NaCl are soluble in water but Ba3(PO4)2 (molar mass = 602 g/mol) is insoluble, the ionic equation for the reaction is: 6 Na+(aq) + 2 PO4-3(aq) + 24 H2O(ℓ) + 3 Ba+2(aq) + 6 Cl-(aq) + 6 H2O(ℓ) → Ba3(PO4)2(s) + 6 Na+(aq) + 6Cl-(aq) + 30 H2O(ℓ) Presenting only the ions that show evidence of a chemical reaction occurring, i.e., the formation of a precipitate (and removing the spectator ions from the equation), the net ionic equation for the observed reaction is: 2 PO4-3(aq) + 3 Ba+2(aq) → Ba3(PO4)2(s) Spectator ions, cations or anions, do not participate in any observable or detectable chemical reaction. Net ionic equation: is an equation that includes only those ions that participate in the observed chemical reaction. From the balanced net ionic equation: 2 mol of PO4-3 react with 3 mol of Ba+2 and form 1 mol of Ba3(PO4)2 or 2 mol Na3PO4.12H2O of react with 3 mol of BaCl2.2H2O and form 1 mol Ba3(PO4)2 In this experiment you will obtain a solid salts mixture of Na3PO4.12H2O and BaCl2.2H2O of unknown composition. The mass of the solid mixture is measured and added to water, the insoluble Ba3(PO4)2 forms. The Ba3(PO4)2 precipitate is collected, via gravity filtration and dried, and its mass is measured. The percent composition of the salt mixture is determines by first testing for the limiting reactant. The limiting reactant is determines from two precipitation tests of the solution:  The solution is tested for an excess of Ba+2 with PO4-3 reagent. The formation of a precipitate indicates the presence of an excess of Ba+2 ion (limited amount of PO4-3 ion). 46   The solution is also tested for an excess of PO4-3 with Ba+2 reagent. The formation of a precipitate indicates the presence of an excess of PO4-3 ion (limited amount of Ba+2 ion). Calculations: The question after collection of data becomes "how do I proceed to determine the percent composition of the salt mixture containing Na3PO4.12H2O and BaCl2.2H2O from the data of precipitation reaction. For illustration consider the following examples: Example (1): A 0.942 g sample of salt mixture is added to water, 0.188 g of Ba3(PO4)2 precipitate forms. Tests reveal that BaCl2.2H2O is the limiting reactant. What is the percent composition of the salt mixture. 2 PO4-3(aq) + 3 Ba+2(aq) → Ba3(PO4)2(s) From the equation: 1 mol Ba3(PO4)2 precipitate requires 3 mol of Ba+2 (therefore 3 mol BaCl2.2H2O). Mass of BaCl2 = 0.188 g Ba3(PO4)2 x 1mol Ba3(PO4)2 x 3 mol BaCl2.2H2O x 244 g BaCl2.2H2O 602 g Ba3(PO4)2 1 mol Ba3(PO4)2 1 mol BaCl2.2H2O = 0.229 g BaCl2 Mass of salt mixture = mass of Na3PO4.12H2O + BaCl2.2H2O 0.942 g = mass of Na3PO4.12H2O + 0.229 g mass of Na3PO4.12H2O = 0.942 g – 0.229 g = 0.713 g The percent of BaCl2.2H2O in salt mixture = (0.229g x 100% = 24.3% 0.942 g the percent of Na3PO4.12H2O in salt mixture = 0.713 g x 100% = 75.7% 0.942 g Example(2): A mixture containing 3.0 g of Na3PO4.12H2O and 4.0 g of BaCl2.2H2O is dissolved in water. A precipitate of Ba3(PO4)2 weighing 2.265 g is produced. Calculate the percent yield of the reaction mol of Na3PO4.12H2O = 3g = 7.89 x 10-3 mol 380 (g/mol) mol of BaCl2.2H2O = 4g = 1.64 x 10-2 mol 244 (g/mol) 47  2 mol PO4-3 3 mol Ba+2 7.89 x 10-3 mol/2 1.64 x 10-2 mol/3 3.95 x 10-3 5.46 x10-3 The limiting reactant is Na3PO4.12H2O mass of Ba3(PO4)2 = 7.89 x 10-3 mol Na3PO4 x 1 mol Ba3(PO4)2 x 602 g Ba3(PO4)2 = 2.37 g 2 mol Na3PO4 1 mol Ba3(PO4)2 percent yield = actual yield x 100% = 2.265 g x 100% = 95.4% theoretical yield 2.37 g Experimental procedure: Two trials are recommended for this experiment. To hasten the analyses, measure the mass of duplicate unknown salt mixtures and simultaneously follow the procedure for each. Label the beakers accordingly for trial (1) and trial (2) to avoid the intermixing of samples and solutions. A) Precipitation of Ba3(PO4)2 : 1. Obtain about 2-3 g of an unknown Na3PO4.12H2O / BaCl2.2H2O salt mixture in a vial from your instructor. 2. Measure the mass of the vial and the salt mixture. Record this mass for trial (1) on the report sheet. 3. Transfer about half of the salt mixture into a labeled 400-ml beaker, reweigh the vial and the salt mixture. Record this mass for trial (1) on the report sheet. 4. Transfer the rest of the salt mixture into another labeled 400-ml beaker, reweigh the empty vial. Record this mass for trial (2) on the report sheet. 5. Add approximately 200 ml of distilled water to the solid salt mixture. 6. Stir the mixture with stirring rod for about 1 minute and then allow the precipitate to settle. Leave the stirring rod in the beaker. 7. Cover the beaker with watchglass and maintain the solution hot (80-90º) on steam bath or with a low flame for about 30 minutes, avoid boiling (This process of heating a precipitate is called digestion of the precipitate. The purpose of the digestion of the precipitate is to obtain larger particles) 48  8. While the precipitate is being kept warm, proceed to set up the gravity filtration apparatus:  Obtain a piece of filter paper and measure its mass.  Fold the filter paper into quarters. Open one fold to form a cone.  Seal the filter paper into the filter funnel with a small amount of distilled water.  Place a 250 ml Erlenmeyer flask below the funnel. 9. After 30 minutes, remove the heat, and allow the precipitate to settle; the solution does not need to cool to room temperature. 10. While the precipitate is setting, heat about 30 ml of distilled water for use as wash water. 11. Once the supernatant has cleared, while still warm, quantitatively transfer the precipitate by first pours most of the liquid without agitating the solid. Then stir and transfer the remaining mixture onto the filter paper. 12. Wash any remaining precipitate onto the filter with several 5-ml portion of hot water 13. Wash the Ba3(PO4)2 precipitate on the filter paper with two additional 5-ml portion of hot water. 14. Remove the filter paper and the precipitate from the filter funnel. 15. Air-dry the precipitate on the filter paper until the next laboratory period or dry in a 110ºC constant temperature drying oven overnight. 16. Determine the combined mass of the precipitate and the filter paper. Record the mass on the data sheet. B) Determining the Limiting Reactant: 1. Test for excess PO4-3: add 2 drops of 0.5 M BaCl2 solution to the 2-3 ml of supernatant in a test tube. If a precipitate forms, the PO4-3 is in excess and Ba+2 is the limiting reactant in the original salt mixture. 2. Test for excess Ba+2: add 2 drops of 0.5 M Na3PO4 solution to the 2-3 ml of supernatant in a test tube. If a precipitate forms, the Ba+2 is in excess and PO4-3 is the limiting reactant in the original salt mixture. 49  EXPERIMENT-Four Pre-laboratory Assignment Limiting Reactant Name: ………………………….. Lab Section: ………………. Date: ………………….. 1. What does the expression "digesting a precipitate" mean? 2. When 0.627 g of Na3PO4.12H2O (molar mass = 380 g/mol) and 0.425 g of BaCl2.2H2O (molar mass = 244 g/mol) are mixed with water forming 500 ml of solution, calculate: a) The mass of the Ba3(PO4)2 (molar mass = 602 g/mol) precipitates b) The mass of the excess reactant 3. 1.2 g sample of BaCl2.2H2O (molar mass = 244 g/mol) react with excess Na3PO4.12H2O. If the mass of Ba3(PO4)2 (molar mass = 602 g/mol) obtained is 0.84 g. Calculate the percent yield of Ba3(PO4)2. 50  EXPERIMENT-Four Report Sheet Limiting Reactant Name: ………………………… Lab Section: ………………. Date: ………………….. DATA SHEET Unknown number: ___________________________ Mass of salt mixture (g) Mass of filter paper (g) Mass of filter paper and Ba3(PO4)2 (g) Mass of Ba3(PO4)2 (g) Limiting reactant in salt mixture: _________________________________ Excess reactant in salt mixture: __________________________________ Calculations: Moles of Ba3(PO4)2 (mol) Moles of the limiting reactant in salt mixture (mol) Mass of the limiting reactant in salt mixture (g) Mass of excess reactant in salt mixture (g) Mass percent of limiting reactant in salt mixture (%) Mass percent of excess reactant in salt mixture (%) Show your calculation for trial (1) 51  EXPERIMENT-Four Post-laboratory Report Limiting Reactant Name: ………………………….. Lab Section: ………………. Date: ………………….. 1. According to your date, calculate the mass of the excess reactant participated in the reaction and how many grams did not 2. Because Ba3(PO4)2 is a very finely divided precipitate, some is lost in the filtering process. As a result is the reported "percent limiting reactant" high or low,..explain 3. Explain the effect of each of the following factors on the actual yield of Ba3(PO4)2: a) Using coarse filter paper instead of one with fine porosity. b) Insufficient washing of the precipitate. c) If the precipitate was not dried completely. 4. Describe how to test your unknown salt mixture for the presence of excess BaCl2.2H2O 52