Lab Manual- Chapter 8
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Purpose
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To determine the limiting reactant in a mixture of two soluble salts. To determine the percent composition of the substances in a salt mixture.
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What are the two factors that affect the yield of products in a chemical reaction?
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1. Percent yield 2. Amounts of the starting reactants.
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Percent Yield
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Actual yield/ Theoretical yield x 100
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Stoichiometry
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By a study of a balanced chemical equation, to determine the moles. 2 H2 + O2 --> 2 H20 (2 moles water for every 1 mole oxygen)
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Limiting Reactant
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The reactant that determines the amount of product generated in a chemical reaction.
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Ionic Equation
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A chemical equation that presents ionic compounds in the form in which they exist in aqueous solutions. Ca ²? + 2Cl? + 2K? + C?O?²? + 3H?O ? CaC?O??H?O + 2Cl? + 2K? + 2H?O
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Spectator Ions
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Cations or Anions that do not participate in any observable or detectable chemical reaction
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Net Ionic Equation
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The Ionic Equation without the spectator ions.
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Part A
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The solid reactant salts form a heterogeneous mixture of an unknown composition. The mass of the solid mixture is measured then added to water. This forms insoluble CaC?O?, this is filtered out, dried and measured.
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Part B
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The percent composition of the salt is determined by first testing for the limiting reactant. The limiting reactant for the formation of solid calcium monohydrate is determined from two precipitation tests of the final mixture from Part A. 1. Mixture is tested for an excess of oxalate ion with limited calcium ion 2. Mixture is tested for an excess of calcium ion with limited oxalate ion.
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A 0.538 sample of the salt mixture is added to water and after drying, a .194 g of CaC?O?*H2O is measured. Tests revealed that K?C?O??H?O was the limiting reagent. What is the percent composition? CaCl??2H?O +K?C?O??H?O ? CaC?O??H?O +2KCl + 2H?O
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1. Find moles of product (146.12) 2. Stochiometric ratio 3. Multiply by Molar mass of K?C?O??H?O (184.24 g/mol) .194 g CaC?O?*H2O * (1 mol/ 146.12) = .0013277 moles CaC?O?*H2O .0013277 moles CaC?O?*H2O * (1 mole K?C?O??H?O/ 1 mole CaC?O?*H2O) = .0013277 mole K?C?O??H?O .0013277 mole K?C?O??H?O * (184.24 g/ 1 mole) = .245 g K?C?O??H?O Percent By Mass: .245/.538 * 100 = 45.5%
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A 0.538 sample of the salt mixture is added to water and after drying, a .194 g of CaC?O?*H2O is measured. Tests revealed that K?C?O??H?O was the limiting reagent. Since K?C?O??H?O was in the limiting reactant, how many grams of the excess reagent were in the mixture? CaCl??2H?O +K?C?O??H?O ? CaC?O??H?O +2KCl + 2H?O
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.538-.245= .293 g CaCl? present in original salt mixture .194 g CaC?O?*H2O (1 mol/ 146.13)= .0013277 mole CaC?O?*H2O .0013277 mole CaC?O?*H2O (1 mole CaCl?/1 mole CaC?O?*H2O) = .0013277 mole CaCl? .0013277 mole CaCl? (147.02 g/ 1 mole)= .195 g CaCl? reacted. (Original salt - reacted salt = grams of excess) .293 - .195= .098
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Procedure
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1. Obtain sample 2. Adjust water pH 3. Mix water and salt 4. Digest the precipitate (heat) 5. Cool for 15 minutes and allow ppt to settle. 6. Set up gravity vacuum w/ filter 7. Dry precipitate. 8. While ppt dries, withdraw supernatant 9. Find the amount reacted. 1. Centrifuge supernatant 2. Test for excess
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What are the reactants and their molar masses in this experiment?
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CaCl? (calcium chloride---147.02) and K?C?O??H?O (potassium oxalate--- 184.02)
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How is the limiting reactant determined in the experiment?
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Save the supernatant from the filtering vacuum. Place into two different test tubes. Centrifuge. Than, test for the reagents. 1. if you put potassium oxalate in and ppt forms-- calcium chloride is excess. 2. if you put calcium chloride in and ppt forms-- potassium oxalate is in excess
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What is the procedure and purpose of "digesting the precipitate"?1
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Purpose: Dissolves the particles so that the filtering process is more efficient. Procedure: Digesting the ppt means that we heat it for approximately 20 minutes.
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Two special steps in the Experimental Procedure are incorporated to reduce the loss of calcium oxalate ppt. Identify these steps and the reasons why.
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1. The ppt is digested. Calcium oxalate (ppt) is reprecipitated onto large particles making the filtering process more efficient. 2. Whatman and Fisherman brand papers are both fine porosity papers, so there is a reduced chance of ppt going through.
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A sample of CaCl??2H?O/K?C?O??H?O solid salt mixture is dissolved in ~150mL of deionized water previously adjusted to a pH that is basic. The precipitate was weighted. 3. Mass of salt mixture .879 4. Mass of filter paper 1.896 5. Mass of filter paper and CaC?O??H?O 2.180 What is the mass of air dried CaC?O??H?O?
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2.180 - 1.896 = .284 g
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A sample of CaCl??2H?O/K?C?O??H?O solid salt mixture is dissolved in ~150mL of deionized water previously adjusted to a pH that is basic. The precipitate was weighted. The limiting reactant was found to be CaCl??2H?O, what is the excess reactant?
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If the limiting reactant was CaCl??2H?O, than the excess must be K?C?O??H?O.
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A sample of CaCl??2H?O/K?C?O??H?O solid salt mixture is dissolved in ~150mL of deionized water previously adjusted to a pH that is basic. The precipitate was weighted. 3. Mass of salt mixture .879 4. Mass of filter paper 1.896 5. Mass of filter paper and CaC?O??H?O 2.180 a.Moles of CaC?O??H?O precipitated? b.Moles of limiting reactant in salt mixture? c.Mass of limiting reactant in salt mixture?
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.284 g CaC?O??H? (1 mol/ 146.13g) = .001943 a. .001943 mole CaC?O??H?O .001943 mole CaC?O??H?O (1mole CaCl??2H?O/1 mole CaC?O??H?O) = .001943 mole CaCl? b. .001943 mole CaCl? .001943 mol CaCl? (147.02 g/ 1 mole) = .28566 g c. .286 g CaCl?
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A sample of CaCl??2H?O/K?C?O??H?O solid salt mixture is dissolved in ~150mL of deionized water previously adjusted to a pH that is basic. The precipitate was weighted. 3. Mass of salt mixture .879 4. Mass of filter paper 1.896 5. Mass of filter paper and CaC?O??H?O 2.180 d. Mass of the excess reactant. e. Percent limiting reactant f. Percent excess reactant
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.879 - .286 = .593 d. .593 g K?C?O??H?O .286/.879= .3254 e. 32.5% .593/.879 = .6746 f. 67.5%
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If the step for digesting the precipitate were omitted, will the reported percent limiting reactant in the mixture be too high, too low, or unaffected?
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Too low. The purpose of digesting the precipitate is to make the filtering process more efficient. If this were to be omitted the amount of product retrieved would be decreased. With a smaller amount of product, there will be an inferred decreased amount in limiting reactant.
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A couple drops of water were accidentally paced on the properly folded filter paper before its mass was measured. However, in part A.6, the precipitate and the paper were dry. As a result of this technique will the reported mass of the limiting reacted be too high, too low, or unaffected?
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Too low, the water adds weight to the paper. When the ppt is dried, the loss in mass will be attributed to the precipitate rather than the paper. A loss in mass of the precipitate means that there will be a loss in the reported limiting reactant.
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Because of the porosity of the filter paper, some of the ppt passes through the filter paper. Will the reported percent of the limiting reactant in the original salt mixture be too high or too low?
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Too low, the loss of your ppt means a dip in the reported percent of the limiting reactant of the original salt.
