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Stoichiometry

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  • 1. High School Chemistry Rapid Learning Series - 12 Rapid Learning Center Chemistry :: Biology :: Physics :: Math Rapid Learning Center Presents … p g Teach Yourself High School Chemistry in 24 Hours 1/32 http://www.RapidLearningCenter.com Stoichiometry HS Ch i t R id Learning Series Chemistry Rapid L i S i Wayne Huang, PhD Kelly Deters, PhD Russell Dahl, PhD Elizabeth James, PhD Rapid Learning Center www.RapidLearningCenter.com/ © Rapid Learning Inc. All rights reserved. © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 1
  • 2. High School Chemistry Rapid Learning Series - 12 Learning Objectives By studying this tutorial you will learn… What stoichiometry is. How dimensional analysis is used to solve stoichiometry problems. How to gather equalities for use in stoichiometry problems. How to use Molarity. How to use the molar volume of a gas at STP f t STP. How to find the limiting reactant. stoichiometry How to find percent yield. 3/32 Concept Map Previous content Chemistry New content Limiting Reactant Studies Percent Yield Can lead to Matter Stoichiometry Undergoes Chemical Equations Shown in Are investigated with Chemical Changes uses Dimensional Di i l Analysis Shows Using Mole Ratio Molar Mass Molarity Molar Volume of a Gas 4/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 2
  • 3. High School Chemistry Rapid Learning Series - 12 Relationships of Compounds in Chemical Equations 5/32 Mole Ratios in Chemical Equations One of the most important pieces of information in a chemical reaction is the mole ratio (shown by the coefficients). and 2 moles of H2O are produced. For every 2 moles of H2… 2 2 2 H2 + O2 2 H2O Coefficient = 1 1 mole of O2 is need to react… 6/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 3
  • 4. High School Chemistry Rapid Learning Series - 12 Using Dimensional Analysis in Stoichiometry 7/32 Definition: Stoichiometry Stoichiometry – Using t e o e at o o t e the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the reaction. CH4+O2 CO2 + 2H2O The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). 8/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 4
  • 5. High School Chemistry Rapid Learning Series - 12 KUDOS & Dimensional Analysis Use the KUDOS method and dimensional analysis to solve stoichiometry problems. K Identify the known. U Identify the unknown. D Identify needed definitions. For dimensional analysis, the “definitions” are the equalities: Ratio from balanced equation (one compound ↔ another compound) Molar mass (grams ↔ moles) Molarity (grams ↔ liters of a solution) Molar volume of a gas (moles ↔ liters of a gas) O Find the output - perform the dimensional analysis. S Substantiate your answer. 9/32 Mole-Mole Problems Example: If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed? 2 H2 + O2 2 H2O From balanced equation: 2 mole H2 1 mole O2 K U 1 mole O2 2 4.2 mole H2 D mole H2 = ________ mole O2 2.1 O 2.1 is a reasonable answer when 4.2 is given. “mole O2” is the correct unit. 2 sig. fig. given 2 sig. fig. in answer. S 10/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 5
  • 6. High School Chemistry Rapid Learning Series - 12 Molar Mass Review Molar mass is needed to convert between grams and moles: 1 Count the number of each type of atom. 2 Find the atomic mass of each atom on the periodic table. 3 Multiply the # of atoms × atomic mass for each atom. 4 Find the sum of all the masses. Example: Find the molar mass for CaBr2 Ca 1 × 40.08 g/mole = 40.08 g/mole Br 2 × 79.91 g/mole = + 159.82 g/mole 199.90 g/mole 1 11/32 2 4 3 1 mole of CaBr2 molecules would have a mass of 199.90 g. Mole-Mass Problems Example: How many grams of AgCl will be precipitated if 0.45 mole AgNO3 is reacted as follows: 2 AgCl + 2 Ca(NO3)2 2 AgNO3 + CaCl2 From balanced equation: 2 mole AgCl 2 mole AgNO3 K U 2 mole AgCl 2 0.45 mole AgNO3 D Molar Mass of AgCl: 1 mole AgCl = 143.35 g mole AgNO3 143.35 g AgCl 1 = ________ g AgCl 65 mole AgCl O 65 is a reasonable answer for grams. “g AgCl” is the correct unit. 2 sig. fig. given 2 sig. fig. in answer. S 12/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 6
  • 7. High School Chemistry Rapid Learning Series - 12 Mass-Mass Problems Example: How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction: 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl From balanced equation: q 2 mole NaOH 1 mole Ba(OH)2 Molar Mass of NaOH: 1 mole NaOH = 40.00 g D Molar Mass of Ba(OH)2: 1 mole Ba(OH)2 = 171.35 g K 14.5 g NaOH 1 mole NaOH 40.00 g NaOH 1 mole Ba(OH)2 171.35 g Ba(OH)2 ( ) 2 mole NaOH 1 mole Ba(OH)2 U 31.1 = ________ g Ba(OH)2 31.1 is a reasonable answer for grams. “g Ba(OH)2” is the correct unit. 3 sig. fig. given 3 sig. fig. in answer. S O 13/32 Moles, Moles Volumes of Solutions and Concentrations 14/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 7
  • 8. High School Chemistry Rapid Learning Series - 12 Definition: Concentration The substance being dissolved in a homogeneous mixture (solution). (solution) Concentration – How much solute is in the solution. Molarity (M) – A concentration unit, t ti it mol/L. 15/32 Definition: Molarity Molarity = moles (solute) L (solution) Molarity gives the number of moles of the solute that are in 1 liter of the solution. 16/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 8
  • 9. High School Chemistry Rapid Learning Series - 12 Using Molarity Molarity is used to convert between moles and liters. Example: If 0.85 moles NaOH are needed and you have a 1.5 M solution, how many liters of the solution do you need? From concentration: 1.5 moles NaOH = 1 L K D U 0.85 mol NaOH 1 L 0.57 = ________ L 1.5 mol NaOH O 0.57 is a reasonable answer for L. “L” is the correct unit. 2 sig. fig. given 2 sig. fig. in answer. S 17/32 Mass-Volume Problems (Solutions) Example: If you need to precipitate 15.7 g of Ba(OH)2, how many liters of 2.50 M NaOH solution is needed? Ba(OH)2 + 2 NaCl 2 NaOH + BaCl2 From balanced equation: q 2 mol NaOH 1 mol Ba(OH)2 Concentration of NaOH: 2.50 mole NaOH = 1 L D Molar Mass of Ba(OH)2: 1 mol Ba(OH)2 = 171.35 g K 15.7 g Ba(OH)2 ( ) 1 mol Ba(OH)2 2 mol NaOH 171.35 g Ba(OH)2 1 mol Ba(OH)2 1 L NaOH 2.50 mol NaOH U 0.0733 = ________ L NaOH 0.0733 is a reasonable answer for L (73.3 mL). “L NaOH” is the correct unit. 3 sig. fig. given 3 sig. fig. in answer. S O 18/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 9
  • 10. High School Chemistry Rapid Learning Series - 12 Molar Volume of a Gas 19/32 Definition: Molar Volume of a Gas Standard Temperature and Pressure (STP) – 1 atm (760 mm Hg) and 273 K (0°C). Molar Volume of a Gas – At STP, 1 mole of any gas = 22.4 liters. 20/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 10
  • 11. High School Chemistry Rapid Learning Series - 12 Mass-Volume Problems (Gases) Example: If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP? 2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g) From balanced equation: q 1 mole Zn 1 mole H2 Molar volume of a gas: 1 mole H2 = 22.4 L D Molar Mass of Zn: 1 mole Zn = 65.39 g K 1.5 g Zn 1 mole Zn 1 mole H2 65.39 g Zn 1 mole Zn 22.4 1 L H2 mole H2 U 0.51 = ________ L H2 0.51 is a reasonable answer for L. “L H2” is the correct unit. 2 sf given 2 sf in answer. S O 21/32 Limiting Reactants 22/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 11
  • 12. High School Chemistry Rapid Learning Series - 12 Planning a Meal You go to the grocery store and you buy 1 package of Brats (5 Brats), 1 package of cheese (16 slices) and 1 p package of hot dog buns (8 buns). g g ( ) If you use all of these… You can make this many… 5 Brats 5 meals 16 slices of cheese 16 meals 8 hot dog buns 8 meals So you have the possibility of making 5, 16 or 8 meals…which is it? You’ll never get the chance to make 8 or 16 meals…you’ll run out of Brats after 5. Once you run out of one component, you have to stop making meals. 23/32 Definition: Limiting Reactant Limiting Reactant – The reactant that runs out first and causes the reaction to stop. In the previous example, the Brats were the limiting reactant—once they were gone you had gone, to stop! Once one of the reactants runs out, the reaction stops…it can’t make any more product. 24/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 12
  • 13. High School Chemistry Rapid Learning Series - 12 Limiting Reactant Problems Example: How many moles of H2O is produced when 2.3 moles of O2 and 2.3 moles of H2 react? 2 H2 + O2 2 H2O From balanced equation: 2 mol H2 2 mol H2O 1 mol O2 2 mol H2O K D U 2.3 mol O2 2 mol H2O 1 2.3 2 3 mol H2 2 mol H2O 2 4.6 = ________ mol H2O mol O2 2.3 = ________ moles H2O mol H2 O You’ll never get to 4.6 moles of H2O because H2 runs out when 2.3 moles H2O is made. 2.3 is reasonable for moles. “moles H2O” is the correct unit. 2 sig. fig. given 2 sig. fig. in answer. S 25/32 Identifying the Limiting Reactant Example: What is the limiting reactant in the last problem? 2 H2 + O2 2 H2O …Calculate and compare the amount of products produced by each reactant . 2.3 mol O2 2 mol H2O 1 2.3 mol H2 2 mol H2O 2 4.6 46 = ________ mol H2O l mol O2 2.3 = ________ mol H2O mol H2 Less product The reactant that produces a lesser amount of product is the limiting reactant. H2 was the reactant that ran out (caused the smallest number of products) therefore, H2 is the limiting reactant. 26/32 Simple Limiting Reagent Finder Mnemonic: Find the moles of each reactant and divide the moles of each reactant by its coefficient. The reactant with the smallest number is the limiting reagent = “Smallest m/c is Limited”. © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 13
  • 14. High School Chemistry Rapid Learning Series - 12 Percent Yield 27/32 Definition: Percent Yield Percent Yield – Compares how much the reaction actually produced (the “actual yield”) to how much the stoichiometry says you should get if things react completely 100% (the “theoretical yield”). % yield = i ld 28/32 actual yield y ×100 theoretical yield % yield is always calculated with masses—not moles! © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 14
  • 15. High School Chemistry Rapid Learning Series - 12 Percent Yield Problem Example: A student works out a stoichiometry problem and determines that they should produce 1.56 g AgCl in the lab. They complete the lab and find that they obtained 1.25 g AgCl. Find the percent yield of the lab. K U D Theoretical yield = 1.56 g AgCl Actual yield = 1.25 g AgCl % yield = ? % yield = % yield = actual yield ×100 theoretical yield O 1.25g AgCl ×100 = 80.1% 1.56 g AgCl 80.1 is a reasonable answer for percent yield. “%” is the correct unit. 3 sig. fig. given 3 sig. fig. in answer. S 29/32 Learning Summary The limiting reactant is the reactant th t runs t t that out first and causes the reaction to stop. Stoichiometry is using the mole ratio from the balanced equation to find information about various compounds. Stoichiometry uses dimensional analysis and various equalities. Percent yield compares the actual yield of a reaction to the theoretical yield from stoichiometry. Molarity is used to form an equality between lit b t moles & liters of a solution. The molar volume of a gas is used to find liters of a gas in stoichiometry. 30/32 © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 15
  • 16. High School Chemistry Rapid Learning Series - 12 Congratulations You have successfully completed the core tutorial Stoichiometry Rapid Learning Center Rapid Learning Center Chemistry :: Biology :: Physics :: Math What’s N t Wh t’ Next … Step 1: Concepts – Core Tutorial (Just Completed) Step 2: Practice – Interactive Problem Drill Step 3: Recap – Super Review Cheat Sheet Go for it! 32/32 http://www.RapidLearningCenter.com © Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com 16