ࡱ> 574 jbjbjj 3]>>>>>>>4h<$`,Rggg,h=>g|ggg=q >>Kq q q gt>>r"@>>>>gq q >>勭Lab: The Oxidation States of Manganese  Introduction The purpose of this lab is to determine the limiting reagent in the formation of a precipitate and to determine the percent composition of the original salt mixture. Two factors limit the yield of products in a chemical reaction. They are: (1) the amounts of starting materials (reactants) and (2) the percent yield of the reaction. Many experimental conditions can be adjusted to increase a reactions yield, but because chemicals react according to fixed mole ratios (stoichiometrically), only a limited amount of product can form from given amounts of starting materials. The reactant that restricts the amount of product is the limiting reagent in that chemical system. To better understand the limiting reagent concept, lets look at the molecular equation for the reaction studied in this experiment: BaCl2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaCl (aq) Since BaCl2 (a dihydrate, BaCl2.2H2O) and Na2SO4 are soluble salts and BaSO4 is a precipitate, the ionic equation for this reaction would be as follows: Ba 2+ (aq) + 2Cl-(aq) + 2Na+(aq) + SO4 2-(aq) -> BaSO4 (s) + 2Na 2+ (aq) + 2Cl-(aq) The net ionic equation is Ba 2+ (aq) + SO4 2- (aq) -> BaSO4 (s) One mole of Ba 2+(from 1 mole of BaCl2.2H2O, 244.2 g/mol) reacts with one mole of SO4 2- (from 1 mole of Na2SO4,, 142.1 g/mol) to produce 1 mole of BaSO4 precipitate (233.4 g/mol), if the reaction proceeds to completion. If , however, only 2.00 g of BaCl2.2H2O and 1.20 g of Na2SO4 are in the reaction vessel, how many moles and grams of BaSO4 would be produced? Since the equation reads in terms of moles, not grams, each reactants number of moles must be determined: 2.00 g BaCl2.2H2O / 244.2 g BaCl2.2H2O = 0.00819 mol BaCl2.2H2O 1.20 g Na2SO4 / 142.1 g Na2SO4 = 0.00844 mol Na2SO4 Since 1.0 mol of Ba 2+ reacts with 1.0 mol SO4 2- , then the 0.00819 mol Ba 2+ in the reaction vessel can only react with 0.00819 mol SO4 2- producing a maximum of 0.00819 mol BaSO4. This consumes all the Ba 2+ in the vessel, i.e., the Ba 2+ is the limiting reagent. This leaves 0.00025 moles of SO4 2- in excess. This sulfate is the excess reagent. Since the limiting reagent is know, the theoretical yield of product, BaSO4, is calculated from the balanced equation. The limiting reagent controls the moles and grams of BaSO4 produced. Since only 0.00819 mol of Ba 2+ were used, then 0.00819 mol of BaSO4 were produced. Therefore, 0.00819 mol BaSO4 x 233.4 g/mol = 1.91 g BaSO4 Thus, 1.91 g of barium sulfate form if the reaction is 100% complete. In this experiment, you will be given a mixture of sodium sulfate, Na2SO4, and barium chloride, BaCl2.2H2O. Its composition is unknown. You will add the mixture to water and precipitate the BaSO4. You will measure the mass and calculate the moles of BaSO4 that precipitate. From the balanced equation you will then be able to calculate the moles and masses of the Na2SO4 and the BaCl2.2H2O in the original mixture. You can then determine the percent composition of the origianl mixture. MATERIALS 1 g of salt mixture dropper bottle of 0.5 M Na2SO4 concentrated HCl dropper bottle of 0.5 M BaCl2.2H2O large beaker ring stand iron ring wire gauze Bunsen burner Procedure PART I: PRECIPITATION OF BaSO4 1. Weigh 1.00 g of unknown salt mixture on weighing paper (0.001g). Transfer it to a 400 mL beaker and add 200 mL of distilled or di-ionized water. Add, using a stirring rod, 1 mL of concentrated HCl (Caution: conc HCl is a severe skin irritant. Flush the affected area with a large amount of water.) Stir the aqueous mixture with the stirring rod for about 1 minute and then allow the precipitate to settle. 2. Cover the beaker with a watch glass and maintain the solution at a temperature between 80oC and 90oC on a steam bath or with a low flame for 40 50 minutes. Allow the precipitate to settle. 3. Gravity filter the precipitate through two thicknesses of pre-weighed filter paper. When all of the precipitate has been transferred to the filter paper, remove the filter paper from the funnel and spread it open on a watch glass to dry overnight. Save the filtrate. PART II: TESTING THE FILTRATE TO DETERMINE THE LIMITING REAGENT 4. Take the filtrate and divide it into 2 50-mL portions . a. Testing for excess Ba 2+: Add 2 drops of 0.5 M Na2SO4 (SO42-) to the 50 mL in beaker I. If a precipitate forms, Ba 2+ is in excess. b. Testing for excess SO4 2-: Add 2 drops of 0.5 M BaCl2.2H2O (Ba 2+) to the 50 mL in beaker II. If a precipitate forms, SO4 2- is in excess. 5. When the filter paper has dried, mass the precipitate. 6. Calculate the grams of barium sulfate formed, the moles of barium sulfate and the percent composition of the original mixture. Data Mass of original mixturemass of 2 pieces of filter papermass of filter paper + precipitateresults of adding Na2SO4result of adding BaCl2.2H2O CONCLUSION: Complete your calculations and write a conclusion. 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