CHM1311 - Lecture(5)

CHM1311 - Lecture(5) - Limiting Reagents (Ch. 4-4) Often...

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Unformatted text preview: Limiting Reagents (Ch. 4-4) Often there is not enough of one reagent to completely consume the other reagent The consumed reagent is the limiting reagent. The leftover reagent is called excess. WARNING: You can not determine the limiting reagent just by looking at the stoichiometric coefficients!! excess Zn: limiting reagent is HCl not enough Zn: limiting reagent is Zn Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) 1 Limiting Reagent Problem: What mass of H2 is produced when 0.50 g Zn is combined with 1.3 g HCl? Which picture does this correspond to? 2 3 4 Conclusion: When 0.50 g Zn is combined with 1.3 g HCl, Zn is the limiting reagent – Zn is entirely consumed and excess HCl remains unreacted. This corresponds to the third flask in the picture below. 1.5 x 10-2 g H2 is formed in the reaction What mass of HCl remains unreacted?? 5 Limiting Reagent Problem: What mass of HCl is left unreacted when 0.50 g Zn is combined with 1.3 g HCl? 6 Summary of Limiting Reagent Problem 0.50 g Zn was combined with 1.3 g HCl Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) From 7.6 x 10-3 mol Zn, 7.6 x 10-3 mol H2 can be produced. (# moles Zn times the stoichiometric factor 1/1). Zn was the limiting reagent 7.6 x 10-3 mol H2 has a mass of 1.5 x 10-2 g Substance HCl Zn H2 Initial Mass 1.3 g 0.50 g 0g Final Mass 0.8 g 0 g (limiting reagent is consumed) 1.5 x 10-2 g 7 Another Example After the reaction of 92.3 g P4 with 286 g Cl2, what are the masses of P4, Cl2, and PCl3? The chemical reaction that occurs is shown below: P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) 8 9 10 11 Summary: reaction of 92.3 g P4 with 286 g Cl2 P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) From 4.03 moles Cl2, 2.69 mol PCl3 can be produced (# moles Cl2 times the stoichiometric factor 4/6) Cl2 is the limiting reagent 2.69 mol PCl3 has a mass of 369 g Substance P4 Cl2 PCl4 Initial Mass 92.3 g 286 g 0g Final Mass 9.0 g (excess reagent) 0 g (limiting reagent is consumed) 369 g 12 To work on: Given the following balanced equation for the reaction of aluminum with aqueous hydrochloric acid which statement below is false? 2Al(s) + 6HCl(aq) → 2AlCl 3 (aq) + 3H 2 (g) 1. For every mole of Al consumed, one mole of AlCl3 is produced. 2. For every mole of H2 produced, 3/2 moles of Al are consumed. 3. For every mole of Al consumed, 3 moles of HCl are consumed. 4. For every mole of HCl consumed, 1/3 moles of AlCl3 and 1/2 moles of H2 are produced.. 13 Concentration Units (Petrucci, Ch. 13 Sec. 13-2) The most common ways of expressing units are: moles solute Molarity (M) = Litres solution moles solute Molality (m) = kilograms solvent moles A Mole fraction (XA) = moles A + moles B + moles C grams solute x 100% Weight percent = grams solution 14 If a solution contains 0.035 moles solute in 2.0 L of water, what is the molarity (assuming the solute occupies a negligible volume)? Concentration Notation: Often instead of saying “the molarity of A is...”, we will write: [A] = Where [A] means “molarity of A” In this problem, [Solute] = 1.8 x 10-2 M 15 KMnO4 is a strong electrolyte: It completely dissociates in water KMnO4(aq) → K+(aq) + MnO4-(aq) (aq) When 0.15 moles KMnO4 is added to enough water to make a 500 mL solution, determine [K+] and [MnO4-] iin n make the solution. the 16 17 When 2 moles of Na2CO3 are added to enough water to make 1 L solution, determine [Na+] and [CO32-]. to Sodium carbonate is ionic and water soluble: Sodium Na2CO3(aq) → 2 Na+(aq) + CO32-(aq) (aq) Na2CO3 18 Preparing Solutions of Specific Preparing Concentration (Ch. 4-3) Concentration moles solute Molarity (M) = Litres solution TWO METHODS: • Weigh out a solid solute and Weigh dissolve in a given quantity of solvent. solvent. • Dilute a concentrated solution Dilute to give one that is less concentrated. concentrated. 19 Dissolve 5.00 g of NiCl2•6 H2O in enough water to in make 250 mL of solution. Calculate the molarity. make 20 For you to work on: 50 mL of solution containing Cl- is titrated to the endpoint with 10 mL of 0.050 M AgNO3. What was the concentration of Cl- in the sample? 1. 2. 3. 4. 5. 1.0x10-3 M 5.0x10-3 M 1.0x10-2 M 5.0x10-2 M 1.0x10-1 M The titration of Cl- with AgNO3. The endpoint is reached when all of the Clhas precipitated and the excess Ag+ reacts with the CrO42- indicator forming a brick-red precipitate. 21 Preparing Solutions by Dilution A solution can also be solution prepared by diluting a concentrated solution to give a solution that is less concentrated. is 22 PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: NaOH and you want 0.50 M NaOH. What do you do? do H2 O Add water to dilute Add the solution the 3.0 M NaOH Concentrated 0.50 M NaOH Dilute 23 How much water is added? 24 25 Preparing Solutions by Preparing Dilution Dilution A shortcut shortcut Minitial • Vinitial = Mfinal • Vfinal 26 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want PROBLEM: 0.50 M NaOH. What do you do? 0.50 M1V1 = M2V2 (3.0 M)(50.0 mL) = (0.50 M)V2 V2 = [(3.0 M)(50.0 mL)]/(0.50 M) = 300 mL 27 Converting Between Converting Concentration Units (Ch. 13-2) Concentration General Strategy Moles of Moles Solute Solute and and Quantity of Solvent Original Original Concentration Units Units New New Concentration Concentration Units Units 28 Laboratory ammonia is 14.8 M NH3 (aq) with a density of 0.8980 g/mL. What is the mole fraction of NH3 in the solution? NH 29 30 31 Calculate the molality of a 5.24 M NaHCO3 solution. (Density of solution = 1.19 g/mL) (Density Molarity (M) = moles solute L solution Molality (m) = moles solute kg solvent 32 33 34 ...
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