Experiment 8: Limiting Reagents PDF

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Limiting Reagents Lab Report, general chemistry...

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Experiment 8: Limiting Reactants 10/3/2018 A. Precipitation of CaC2O4•H2O from the Salt Mixture Trial 1 Trial 2 1. Mass of beaker (g) 143.0837 103.5681 2. Mass of beaker and salt mixture (g) 144.0857 104.5703 3. Mass of salt mixture (g) 1.0020 1.0022 4. Mass of filter paper (g) 0.4365 0 .4355 5. Mass of filter paper and product after air-dried or oven-dried (g) 0.7378 0.7509 6. Mass of dried product (g) 0.3013 0.3154 7. Formula of dried product CaCl2O4* H2O B. Determination of Limiting Reactant 1. Limiting reactant in salt mixture (write complete formula) CaCl2* 2H2O 2. Excess reactant in salt mixture (write complete formula) K2C2O4*H2O Data Analysis 1. Moles of CaC2O4•H2O (or CaC2O4) precipitated (mol) 0.002062 0.002159 2. Moles of limiting reactant in salt mixture (mol) 0.002062 0.002159 • formula of limiting hydrate 3. Mass of limiting reactant in salt mixture (g) 0.3032 0.3174 4. Mass of excess reactant in salt mixture (g) 0.6988 0.6848 • formula of excess hydrate 5. Percent limiting reactant in salt mixture (%) 30.25 31.67 6. Percent excess reactant in salt mixture (%) 69.75 68.33 7. Mass of excess reactant that reacted (g) 0.3799 0.3978 8. Mass of excess reactant, unreacted (g) 0.3189 0.2870 Laboratory Questions 1. Part A.2. If the step for digesting the precipitate were omitted, will the reported “percent limiting reactant” in the salt mixture be too high, too low, or unaffected? Explain. 1. The Percent limiting reactant will be too low because the point of digesting the precipitate is to make filtering the solution more efficient. This would cause a smaller amount of product and there will be an infrared decreased amount in the limiting reactant. b. Part A.3. A couple of drops of water were accidentally placed on the properly folded filter paper before its mass was measured. However, in Part A.6, the CaC2O4•H2O precipitate and the filter paper were dry. As a result of this sloppy technique, will the mass of the limiting reactant be reported too high, too low, or remain unaffected? Explain. 1. The mass of the limiting reactant would be too low. The water adds mass to the filter and when it dries and is measured with the precipitate it would make it appear there was more mass lost in precipitate than should have been. b. Part A.5. Because of the porosity of the filter paper some of the CaC2O4•H2O precipitate passes through the filter paper. Will the reported percent of the limiting reactant in the original salt mixture be reported too high or too low? Explain.

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1. The reported percent of the limiting reactant in the original salt mixture will be reported to low as the loss of precipitate means a lower reported percent of limiting reactant of original salt. Part A.5. Excessive quantities of wash water are added to the CaC2O4•H2O precipitate. Will the mass of the CaC2O4•H2O be reported too high, too low, or remain unaffected? Explain. 1. The mass of the CaC2O4•H2O be reported too low for excessive amounts of water would result in a larger mass loss of precipitate. Part A.6. The CaC2O4•H2O precipitate is not completely air-dried when its mass is determined. Will the reported mass of the limiting reactant in the original salt mixture be reported too high or too low? Explain. 1. The reported mass of the limiting reactant will be too high for the mass of water would increase total mass of the filter paper and salt making it appear that the limiting reactant is higher percent than it is. Part A.6, 7. The drying oven, although thought (and assumed) to be set at 125°C, had an inside temperature of 84°C. How will this error affect the reported percent by mass of the limiting reactant in the salt mixture . . . too high, too low, or unaffected? Explain. 1. The mass of the limiting reactant in the salt mixture will be too high because the mass could still change before it is completely air dried. To prevent this the sample should be dried completely before recording mass. Part A.4 and Part B. In a hurry to complete the experiment, Anna withdrew two volumes of solution from Part A.2 before the precipitate had settled. As a result, what dilemma might she have encountered in Part B? Explain. 1. Anna would not be able to calculate what’s in excess because the solution would contain both Ca²⁺ and C₂O₄ ²⁻ in both volumes. Part B.2. A reagent bottle on the shelf labeled 0.5 M NaCl was used in place of the 0.5 M CaCl2. AssumingC2O42- to be in excess, what would be observed as a result of using this wrong reagent in this test? Explain. 1. No precipitate will form if 0.5M NaCl instead of 0.5 M CaCl2 as C2O42- forms soluble solution with Na+ but forms precipitate with Ca2+...