Stoichiometry of Precipitation Reaction

Stoichiometry of a rush Reaction March 20,2013 Amber McCollum Introduction Stoichiometry is a branch of interpersonal chemistry that deals with the quantitative relationships that exist among the reactants and out arrogates in chemical reactions To predict the amount of product produced in a precipitation reaction using stoichiometry, accurately account the reactants and products of the reaction, determine the actual rec in all vs. the theoretical comport and to calculate the shargon go. The equation that depart be used is Ba(NO3)2 (aq) + CuSO4 (aq) BaSO4 (s) + Cu(NO3)2 (aq) Method 1. Gather materials undeniable for taste which included a. petite test tube with lip b. Large beaker c. Small graduated cylinder d. Large graduated cylinder e. One 9in heave f. Citric acid g. Sodium bicarbonate h. Sodium chloride 2. To start the experiment * Na2CO3(aq) + CaCl2. 2H2O(aq) a CaCO3(s) + 2NaCl(aq) + 2H2O * Put on your goggles. * Weigh out 1. 0 g of CaCl22H2O and put it into the 100- mL beaker. Add 25 mL of distilled peeing and stir to exploit the calcium chloride solution. Use only distilled water since tap water may have impurities that interfere with the experiment.. Use stoichiometry to determine how much Na2CO3 you will gather up for a full reaction. Weigh the calculated amount of Na2CO3 and put it in a small paper cup. Add 25 mL of distilled water and stir to make a atomic number 11 carbonate solution. * swarm the sodium carbonate solution from the paper cup into the beaker with the calcium chloride solution. A fall down of calcium carbonate will form instantly. * Use the following instructions to set up a filtproportionn assembly. * Swirl the contents of the beaker to dislodge any(prenominal) precipitate from the sides. Then, while holding the separate out paper in send and open, slowly pour the content of the beaker into the filter paperlined funnel.Be careful to not permit the solution overflow the level of the filter paper while pouring. * banner out 2 to 5 mL of distilled water into the graduated cylinder. Pour this down the sides of the beaker, swirl, and pour into the filter paper-lined funnel. * after all the liquid has course from the funnel, lay the filter paper containing the precipitate on folded layers of paper towels and put this someplace where it will not be disturbed while the filter paper and its contents air-dry. Depending upon the humidity in your area this might ready several hours or days. When the filter paper and the precipitated calcium carbonate are completely dry weigh them, subtract the original weight of the forsake filter paper, and record the net weight of the calcium carbonate. This is your actual yield of calcium carbonate. * Now calculate the percent yield, using your theoretical yield and actual yield. Make sure to show all stoichiometric calculations and all data in your lab report. Calculations cadence 1 metamorphose 2 g of Ba(NO3)2 to mols of Ba(NO3)2 2 g Ba(NO3)2 x 1 mol Ba(N O3)2 = 0. 00765 groins Ba(NO3)2 261. 4 g Ba(NO3)2 Step 2 Consider the mol ratios of Ba(NO3)2 and CuSO4.The equation tells us that for 1 mole of Ba(NO3)2 we wishing 1 mole of CuSO4. Thus, since the mole ratio is 11, if we have 0. 00765 moles of Ba(NO3)2 we will need 0. 00765 moles of CuSO4. Step 3 Convert moles of CuSO4 to grams of CuSO4. 0. 00765 moles CuSO4 x 159. 6 g CuSO4 = 1. 22 g CuSO4 1 mole CuSO4 This means that we need 1. 22 g of CuSO4 to fully react with 2 g of Ba(NO3)2. Step 4 How much BaSO4 can we brook? The mole ratio between Ba(NO3)2 and BaSO4(s) is also 11. That means if we have 0. 00765 moles of Ba(NO3)2 we will also get 0. 00765 moles of BaSO4(s).Step 5 Convert the moles of BaSO4 to grams of BaSO4. 0. 00765 moles BaSO4 x 233. 4 g BaSO4 = 1. 79 g BaSO4 1mole BaSO4 Step 6 Double check our results by calculating the amount of Cu(NO3)2 (aq). We dont really need to live on the amount of Cu(NO3)2 (aq) for the experiment, but it helps us double check our other results. Since we admit that the sum up mass of reactants must equal the total mass of products, we suppose 0. 00765 moles Cu(NO3)2 x 187. 55 g Cu(NO3)2 = 1. 43 g Cu(NO3)2 1 mole Cu(NO3)2 Thus, 2 g Ba(NO3)2 positive 1. 22 grams CuSO4, yields 1. 79 g BaSO4. plus 1. 43 g Cu(NO3)2.We can verify our results by comparing the total mass of reactants, 3. 22 g, with the total mass of products, also 3. 22 g. This tells us that all our calculations are correct and we can confidently use them. Step 7 Calculate the theoretical yield. From previous calculations we know that we started with 2 grams of Ba(NO3)2, and need 1. 22 grams of CuSO4 to complete the reaction from which we can expect a yield of 1. 79 grams of BaSO4. Yet this is only a theoretical yield, for we should realistically expect a little less due to expected experimental demerit such as some BaSO4 being lost as it passed with the filter paper.Step 8 Determine the actual yield and percent yield. After the reaction is completed and the precipitate has formed, we need to filter and dry the precipitate before we can weigh it. If we assume that after drying we have 1. 65 grams of BaSO4, then The theoretical yield is 1. 79 grams of BaSO4. The actual yield is 1. 65 grams of BaSO4. The percent yield is 1. 65 g/ 1. 79 g x 100 = 92. 2%. Conclusion After the testing each know and unacknowledged of the experiment, finding the ratio of the substances wasnt very hard. The percentage of the unknown was 85. 8 %.