Stoichiometry, which is the relationship between the amount of reactants used and the amount of product produced in a reaction, is used by scientists everyday when dealing with chemical reactions. In this lab, the limiting reactant and the the excess reactant was found, which was used to find the theoretical mass of the precipitate in the reaction. Lastly, the theoretical mass and the actual mass of the precipitate, which was obtained from a sodium carbonate and calcium chloride reaction, were compared in order to calculate the percentage yield.
Just like scientists, we calculated these amounts using different stoichiometry calculations HypothesisThe Na2CO3 solution and CaCl2 solution react, creating a precipitate CaCO3 and salt (Na2CO3). This reaction will have a limiting reactant and an excess reactant since its difficult to measure and mix the same amount for the two reactants. The solid product should also weigh around 1.25 grams since its theoretical mass is 1.
25 grams. Experimental Procedure In this lab, 25 ml of Na2CO3 solution and CaCl2 solution was measured. The molarity of both solutions was given. The weight of the filter paper was also measured. After measuring the materials, a glass funnel stuffed with the filter paper was placed on a buret stand, and a 250ml beaker was placed under it to capture the “waste” of the reaction. The two reactants were mixed in another 250ml beaker. The product was left to sit in the beaker for a few minutes for the reaction to complete.
The white product of the reaction was then poured into the funnel and left for another few minutes to filter. Once the filtration was complete, the solid on the filter paper was left to dry until the end of the school day. The dry filter paper and product were measured. Data & Results The results of this lab is shown in the table below.Experimental Data and Measurements Mass (g)Weight of the filter paper1.
03 gWeight of the filter paper with NaCl2.45 gWeight of NaCl1.42 gUsing the molarity and the volume of the two reactants, the number of moles of Na2CO3 and CaCl2 were calculated:moles of Na2CO3 =Concentration volume moles of CaCl2=concentrationvolume =0.
7m/L0.025L =0.5m/L0.
025L = 0.0175mol =0.0125mol =0.
018mol =0.013molThen, the number of moles were used to help solve for the limiting reactant and the excess reactant:0.175mol Na2CO31mol CaCO31 mol Na2CO3100.1g1 mol CaCO3=17.52g CaCO30.
0125mol CaCl21 mol CaCO31 mol CaCl2100.1g1 mol CaCO3=1.25g CaCO3Since less amount of CaCO3 could be created using CaCl2, CaCl2 was the limiting reactant and Na2CO3 was the excess reactant. Therefore, 1.25 grams of CaCO3 precipitate could be produced in this reaction.
Lastly, the percentage yield of the theoretical mass and the actual mass of the precipitate was calculated: percentage yield =mass of product obtained mass of product expected =1.42g1.25g =113.6%Observations In this lab, when the precipitate got filtered, then dried, its mass was calculated to be 1.
42 grams. This was very close to the theoretical mass that was calculated to be 1.25 grams. When the percentage yield of this reaction was calculated, there was only a 13.6% difference between the actual mass of the precipitate and its theoretical mass which is within the reasonable margin of error.
We were also able to successfully calculate the limiting reactant. Based on our calculations, the limiting reactant was the calcium chloride solution. The objectives of this lab were achieved because we were able to observe the reaction, calculate the limiting and the excess reactant, and theoretical mass of the precipitate. And lastly, we were able to successfully compare the theoretical mass and the actual mass of the precipitate using percentage yield. To improve our results and decrease the 13.6% error in this lab, we could measure the two reactants more accurately, paying more attention to the meniscus of the liquid. In order to measure the complete amount of the precipitate, we could try to leave no BaCO3 in the beaker when pouring the precipitate on the filter paper by rinsing the beaker with more water. Conclusion Follow- Up Questions 1.
If I were to dry the filtered solution, NaCl, I would only be left with solid NaCl. Salt is soluble in water and when it dissolves, its molecule separates into Na+ and Cl-. It separates because the two atoms get pulled apart, then surrounded by H2O molecules which prevents them from forming ionic bonds. If the water evaporates away, the Na+ and the Cl- atoms will be able to form ionic bonds again, turning back into solid NaCl, table salt. 2. The theoretical volume of Na2CO3 that would result in no excess reactant would be about 0.
018L as proved in the calculation below: number of moles concentration =volume To check: # of moles=0.70.0178? =0.
0125 mol0.0125mol0.7mol/L=0.
0178L Na2CO3 3. a) The balanced equation that describes the of BaCl2 and Na2CO3 is: BaCl2 + Na2CO3 ? BaCO3 + 2NaCl b) To find the limiting reactant, I first found the moles of the two reactants, then I used the mole bridge to find the mass of the product for each reactant: BaCl2: # of mol=0.5M0.05L Na2CO3: # of mol=0.
75M0.075L =0.025 mol =0.056 mol0.025mol BaCl21mol BaCO31 mol BaCl2197.
3g1 mol BaCO3=4.93g BaCO30.056mol Na2CO31mol BaCO31 mol Na2CO3197.3g1 mol BaCO3=11.05g BaCO3 BaCl2 is the limiting reactant. c) The theoretical mass of BaCO3 that should form is 4.93 grams since the limiting reactant produces 4,93 grams of BaCO3. d) To find the actual mass of BaCO3 formed, I multiplied the percentage (82%) by the theoretical mass of BaCO3.
Then, I calculated the percentage yield to check my answer: 0.824.93=4.043grams % yield =mass of product obtained mass of product expected = 4.04g4.93g100 =82% ?
