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Second Exam: Friday February 15 Chapters 3 and 4. Please note that there is a class at 1 pm so you will need to finish by 12:55 pm. Electronic Homework due R by 11:30 pm Office hours this week: T 2-3 pm R 1-2 pm SL 130

Solving Limiting Reactant Problems in Solution - Precipitation Problem - I Problem: Lead has been used as a glaze for pottery for years, and can be a problem if not fired properly in an oven, and is leachable from the pottery. Vinegar is used in leaching tests, followed by Lead precipitated as a sulfide. If 257.8 ml of a 0.0468 M solution of Lead nitrate is added to 156.00 ml of a 0.095 M solution of Sodium sulfide, what mass of solid Lead Sulfide will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Writing the balanced equation: Pb(NO 3 ) 2 (aq) + Na 2 S (aq) 2 NaNO 3 (aq) + PbS (s)

Volume (L) of Pb(NO 3 ) 2 solution Mass (g) of PbS Amount (mol) of Pb(NO 3 ) 2 Volume (L) of Na 2 S solution Amount (mol) of Na 2 S Amount (mol) of PbS Amount (mol) of PbS Multiply by M (mol/L) Multiply by M (mol/L) Molar Ratio Choose the lower number of PbS and multiply by M (g/mol)

Volume (L) of Pb(NO 3 ) 2 solution Volume (L) of Na 2 S solution Amount (mol) of Pb(NO 3 ) 2 Amount (mol) of Na 2 S Amount (mol) of PbS Mass (g) of PbS Multiply by M (mol/L) Multiply by M (mol/L) Molar Ratio Divide by equation coefficient Divide by equation coefficient Smallest

Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO 3 ) 2 = V x M = 0.2578 L x (0.0468 Mol/L) = = Moles Na 2 S = V x M = 0.156 L x (0.095 Mol/L) =

Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO 3 ) 2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb +2 Moles Na 2 S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Calculation of product yield: Therefore Lead Nitrate is the Limiting Reactant!

Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO 3 ) 2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb +2 Moles Na 2 S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: Moles PbS = 0.012065 Mol Pb +2 x = 0.012065 Mol Pb +2 1 mol PbS 1 mol Pb(NO 3 ) 2

Solving Limiting Reactant Problems in Solution - Precipitation Problem - II Moles Pb(NO 3 ) 2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb +2 Moles Na 2 S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: Moles PbS = 0.012065 Mol Pb +2 x = 0.012065 Mol Pb +2 1 mol PbS 1 mol Pb(NO 3 ) 2 0.012065 Mol Pb +2 = 0.012065 Mol PbS 0.012065 Mol PbS x = 2.89 g PbS 239.3 g PbS 1 Mol PbS

Figure 4.21A: Gravimetric analysis for barium: solution is poured. Photo courtesy of James Scherer.

Figure 4.21B: Gravimetric analysis for barium: solution filtered.

Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: moles N 2 = = 85.90 g N 2 28.02 g N 2 1 mole N 2 moles H 2 = = 21.66 g H 2 2.016 g H 2 1 mole H 2

Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: moles N 2 = = 3.066 mol N 2 85.90 g N 2 28.02 g N 2 1 mole N 2 moles H 2 = = 10.74 mol H 2 21.66 g H 2 2.016 g H 2 1 mole H 2

Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: 3.066 mol N 2 2 mol NH 3 1 mol N 2 10.74 mol H 2 mol NH 3 3 mol H 2 = 6.132 mol NH 3 = 7.16 mol NH 3 N 2 is Limiting!

Percent Yield of a reaction: Actual Yield x 100 Theortetical Yield

Percent Yield/Limiting Reactant Problem - II Solution Cont. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Since Nitrogen is limiting, the theoretical yield of ammonia is: 6.132 mol NH 3 x = 104.427 g NH 3 (Theoretical Yield) 17.03 g NH 3 1 mol NH 3 Percent Yield = x 100% Percent Yield = x 100% = 94.49 % Actual Yield Theoretical Yield 98.67 g NH 3 104.427 g NH 3

CaCO 3(s) + 2 HCl (aq)  CaCl 2(aq) + CO 2(g) + H 2 O (l) 2 g10 mL 0.75 M Which is limiting? 2 g CaCO 3 x 1 mol CaCl 2= 0.01 mol CaCl 2 100 g CaCO 3 1 mol CaCO 3 0.01 L HCl x 0.75 mol HCl x 1 mol CaCl 2 = L HCl 2 mol HCl 0.004 mol CaCl 2 How many g of CaCO 3 remain? What is the [Cl - ] after the reaction?

Fig. 3.14

Figure 4.22A: Titration of an unknown amount of HCl with NaOH (#1). Photo courtesy of American Color.

Figure 4.22B: Titration of an unknown amount of HCl with NaOH (#2). Photo courtesy of American Color.

Figure 4.22C: Titration of an unknown amount of HCl with NaOH (#3). Photo courtesy of American Color.

Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO 4 and has a molecular mass of 158.04 g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO 4 into sufficient water to make 250.00 ml of solution. What is the molarity of this diluted solution? 1.58 g KMnO 4 x = 1 mole KMnO 4 158.04 g KMnO 4

Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO 4 and has a molecular mass of 158.04 g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO 4 into sufficient water to make 250.00 ml of solution. What is the molarity of this diluted solution? 1.58 g KMnO 4 x = 0.0100 moles KMnO 4 1 mole KMnO 4 158.04 g KMnO 4

Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO 4 and has a molecular mass of 158.04 g / mole Problem: Prepare a solution by dissolving 1.58 grams of KMnO 4 into sufficient water to make 250.00 ml of solution. What is the molarity of this diluted solution? 1.58 g KMnO 4 x = 0.0100 moles KMnO 4 1 mole KMnO 4 158.04 g KMnO 4 Molarity = = 0.0400 M 0.0100 moles KMnO 4 0.250 liters Molarity of K + ion = [K + ] ion = [MnO 4 - ] ion = 0.0400 M

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