Experiment 9 (2)

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Apr 3, 2024

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9 Experiment 9 Limiting the Reaction Learning Objectives: Upon completion of this experiment, students will: 1. (CLO3). Analyze evidence to decide if generalizations or conclusions based on the obtained data are warranted 2. (CLO4). Interpret and utilize mathematical formulas while solving problems 3. (MLO) Demonstrate an understanding of stoichiometry by calculating theoretical yield, actual yield, percent yield and limiting reagent. 4. Experience using gravity filtration and gravimetric analysis. Discussion: The limiting reactant of a reaction is the chemical species which is the first to be completely consumed in the reaction. It is the reactant that limits the amount of products which can be made. As an example, consider the construction of a car from 10 wheels and 2 frames. Only 2 cars can be made with two wheels left over or in excess, making the frames the limiting reactant. You will prepare a salt mixture of strontium chloride hexahydrate, SrCl 2 ·6H 2 O, and sodium phosphate dodecahydrate, Na 3 PO 4 ·12H 2 O, and allow them to react in approximately 150 mL of deionized water. Both of these salts are soluble in water but the product, strontium phosphate is not. In the reaction mixture, one of the hydrated salts will limit the amount of strontium phosphate is produced. Once the precipitation reaction is complete, the precipitate is filtered, dried, and weighed. The precipitate is captured in the filter paper, and the liquid that passes through the filter paper, called the filtrate, is collected and tested with very concentrated solutions of the reactants. The extent of the reaction is being tested. For example, if we go back to the production of the cars, we made two cars with 2 wheels left over. If we buy more wheels, we cannot make more cars because we have even more of the wheels in excess; however if we purchase more frames, we could begin to make more cars with the leftover wheels. In this reaction, the reactant that is in excess is determined by testing the filtrate with more of the reactants. The filtrate will clearly show a precipitate if there is excess reactant present with which the testing reactant can react. If strontium chloride is in excess, adding strontium chloride will not show a precipitate, but by adding the reactant, sodium phosphate, which Page 1

reacts with strontium chloride, the resulting solution will show a precipitate of strontium phosphate. In this reaction, PO 4 3- ions react with Sr 2+ ions in excess producing a precipitate. The same is true if strontium ions are added to a solution that contains phosphate ions. If only one reaction shows a precipitate, the limiting reactant can be determined. If both tubes show cloudiness, it means that the reaction did not go to completion. If neither reaction shows a precipitate, then both reactants were added so that both react completely. Before the precipitate can be filtered, the precipitate particles must be large enough that they are trapped by the filter paper. Precipitation begins by a process called nucleation in which tiny particles of the precipitate form. These particles are too small to be trapped by the filter paper. The particle’s size of the precipitate can be increased through digestion, a gentle heating process. Nucleation is favored in saturated solutions, and particle growth is favored in unsaturated solutions. A saturated solution is one that holds as much solute as possible at the given temperature. An unsaturated solution holds less than the maximum amount of solute. When water is added to the solid salt mixture the resulting solution is relatively saturated, which is not the best situation for creating large precipitate particles that are easy to filter. In order to encourage particle growth the saturation of the solution is decreased by gentle heating. After the strontium phosphate is filtered, dried, and weighed, the starting mass of the limiting reactant salt can be determined. The calculation of quantities of reactants used and products created can be determined using the process of calculations called stoichiometry. Stoichiometry relates the mass of each substance to it number of moles and the moles of the reactants to the moles of products using the balanced chemical reaction. This process can be used under any situation where a chemical reaction is known and the quantities measured. Stoichiometry can be shown as mass A ÷ molar mass of A = moles A ; mass A × 1 mol MM ( g ) = mol A moles A × coefficient of B ∈ balanced reaction coefficient of A ∈ balanced reaction = theoretical moles B moles B × molar mass of B = mass B ; mol B × MM ( g ) 1 mole = theoretical yield or mass B The first step in any stoichiometry problem is to determine the balanced chemical reaction. In most of the examples you will be given the reaction or the reactants and must determine the Page 2

overall balanced reaction. In this problem, the two ionic compounds react in a double replacement reaction and the products are easily predicted from the reactants. The mass of a single reactant or in this case the mass of both reactants are then measured. The molar mass of the reactants can be calculated from the atomic masses, thus allowing us to calculate the moles of each reactant. For example if we wish to determine the moles of CaCl 2 present in 10.00 g of the salt, then 10.00 g CaCl 2 × 1 molCaCl 2 110.984 gCaC l 2 = 0.09010 mol CaCl If CaCl 2 is reacted with Na 2 SO 4, , the balanced reaction is CaCl 2 (aq) + Na 2 SO 4 (aq) → 2 NaCl (aq) + CaSO 4 (solid precipitate) Then the mass of the precipitate can be determined as 0.09010 mol CaCl 2 × 1 molCaS O 4 1 molCaCl 2 = 0.09010 mol CaSO 4 And the theoretical mass of the precipitate calculated using the molar mass of the calcium sulfate as 0.09010 mol CaSO 4 × 136.141 gCa SO 4 1 molCa SO 4 = 12.27 g CaSO 4 If a sample of mixed Na 2 SO 4 and CaCl 2 with a mass of 1.017 g was dissolved, digested, and filtered, can we determine the percent of the original sample that was the limiting reactant? First we have to determine which substance was limiting by the reagent test. Precipitation tests on the supernatant determined that Na 2 SO 4 was the limiting reactant. The precipitate was collected on a 5.000 g piece of filter paper. Once dried, the filter paper and precipitate has a combined mass of 5.198 g. The mass of calcium sulfate is 0.198 g, or 1.45 × 10 -3 mol CaSO 4. The precipitation test on the supernatant lets us know that the sodium sulfate is limiting, which means that all of the sulfate is present in the precipitated calcium sulfate. This means that the moles of Na 2 SO 4 can be determined from the moles of precipitate. Note that the moles of calcium cannot be determined from the moles of calcium sulfate, since the calcium chloride was in excess, there is excess calcium ions in the supernatant. Starting from the moles of calcium sulfate precipitate, we calculate the moles of sodium sulfate in the original salt mixture: 0.198 g CaSO 4 × 1 molCaSO 4 136.141 gCaS O 4 × 1 mol Na 2 SO 4 1 molCaS O 4 = 0.00145 mol Na 2 SO 4 0.00145 mol Na 2 SO 4 × 142.042 g Na 2 SO 4 1 molNa 2 S O 4 = 0.206 g Na 2 SO 4 From this, the mass percent of limiting reactant can be calculated: Page 3

0.206 g Na 2 SO 4 1.017 g sample × 100 = 20.3% Na 2 SO 4 The mass of the in excess reactant can be determined by difference. If 1.017 g of the mixed salts contained 0.206 g was the limiting reactant, then 0.811 g of CaCl 2 which is 79.7% of the mixture. Procedure: Label two beakers Beaker I and II. In the beaker labeled Beaker I, separately weigh 1.0 g of Na 3 PO 4 ·12H 2 O and 1.5 g of SrCl 2 ·6H 2 O, record the exact mass of each salt added to the beaker. In the beaker labeled Beaker II, weigh approximately 2 g of the solid Na 3 PO 4 ·12H 2 O - SrCl 2 ·6H 2 O mixture provided , record the exact mass of the mixture added to the beaker. The procedure for each beaker is the same at this point. Repeat the following procedure for both Beaker I and Beaker II. Add approximately 150 mL of deionized water to the beaker while stirring, continue stirring for approximately 3 minutes. Leave the stir rod in the beaker, cover with a watch glass, and allow the undissolved salts or precipitate to settle to the bottom of the beaker. Digest the precipitate for 15 minutes using a hot plate on low heat or place the beaker on the ring stand over a cool blue flame on a Bunsen burner. Do not let the mixture come to a boil. Remove the heat from the beaker if the mixture begins to boil. After 15 minutes, remove the beaker containing the sample from the heat and let it stand until the mixture and beaker is cool enough to be safely handled. Heat about 100 mL of deionized water to between 70 - 80°C. This water will be used for washing the precipitate out of the beakers and off the stir rods. This water will also be used to rinse the precipitate in the funnel. While the sample cools and the wash water is heating, set up the filtration apparatus. The filtration can be done by gravity filtration or by vacuum filtration. Obtain one sheet of fine porosity filter paper for each beaker. Using pencil, label the filter paper with your initials and sample number. Record the mass of each dry filter paper. Filter the sample and set the filtrate aside for testing. Transfer all of the precipitate into the funnel, rinse the precipitate with the warmed wash water. DO NOT throw away the filtrate. Place the filter paper on a clean watch glass to dry. The sample can be dried in a drying oven to a constant mass of + 0.01 grams or placed in a safe place until the next laboratory period. Allow the filtrate to settle and carefully decant 5 mL of the filtrate into two test tubes, labeled SrCl 2 and Na 3 PO 4 . If the filtrate is still slightly cloudy, centrifuge the samples or allow the precipitate to settle to the bottom of the test tube . Add 5 drops of 1M SrCl 2 test reagent to tube labeled SrCl 2 to test for the presence of phosphate ion in excess and 5 drops of 1 M Na 3 PO 4 reagent to test tube labeled Na 3 PO 4 to test for excess strontium ions. You should see an obvious precipitation reaction in only one of the tubes. Dispose of the filtrate in the liquid waste container and clean Page 4

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