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Lesson 11 Stoichiometry

Lesson 11 Stoichiometry. 11.1 – How Much??. 11.2 – Mole Ratios. 11.3 – Excess and Limiting Reactants. 11.4 – Percent Yield. 11.5 – Molarity. Anything in black letters = write it in your notes (‘knowts’). 11.1 – How Much??. Stoichiometry -. Quantitative study of chemical rxns.

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Lesson 11 Stoichiometry

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  1. Lesson 11Stoichiometry 11.1 – How Much?? 11.2 – Mole Ratios 11.3 – Excess and Limiting Reactants 11.4 – Percent Yield 11.5 – Molarity Anything in black letters = write it in your notes (‘knowts’)

  2. 11.1 – How Much?? Stoichiometry - Quantitative study of chemical rxns. In this lesson we will be asking ‘how much?’

  3. The coefficients of a balanced chemical equation represent the number of moles that are reacting or produced. 2H2 + O2 2H2O

  4. 2 ( ) + 1 ( ) 2 ( ) 6.02  1023 molecules H2 6.02  1023 molecules O2 6.02  1023 molecules H3O 2H2 + O22H2O 2H2 molecules +1O2 molecule 2H2O molecules It is not practical to talk about single molecules; instead use a larger number of molecules… It is even more practical to talk about moles… 2mol of H2+ 1mol of O2→ 2mol of H2O

  5. The bike example… For simplicity, say a bike requires only 1 frame and 2 wheels. 2wheels + 1frame  1bike What are the coefficients here? What do they tell you?

  6. + → 2wheels + 1frame  1bike How many frames would be needed to ‘react’ completely with 20 wheels? How many bikes could be produced from 4 wheels and 560 frames? What is the limiting reactant here?

  7. +  2wheels + 1frame  1bike How many bikes would be produced from 23.7 kg of wheels and 80.1 kg of frames? What is needed in order to solve the above question? Always convert to a number of things (mol) first!

  8. Complete each row in the chart…

  9. 2Mg + O2 2MgO Complete each row in the chart using the given amounts

  10. 11.2 – Mole Ratios A mole ratio comes from the coefficients of a balanced chemical equation. Mole ratios are used to compare the amount of mol of one substance to another.

  11. 1 mol N2 3 mol H2 2 mol NH3 1 mol N2 3 mol H2 2 mol NH3 3 mol H2 1 mol N2 2 mol NH3 1 mol N2 2 mol NH3 3 mol H2 N2(g) + 3H2(g)  2NH3(g) Write the three mole ratios that can be written from this balanced equation… These are equivalent ratios, just upside down…

  12. N2(g) + 3H2(g)  2NH3(g) How many moles of NH3 are produced when 1.0 mol of N2 reacts with excess H2? How many moles of NH3 are produced when 0.60 mol of N2 reacts with excess H2? 2 mol NH3  = 1.2 mol NH3 0.60 mol N2 1 mol N2

  13. N2(g) + 3H2(g)  2NH3(g) Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. = 31 g NH3

  14. 4P(s) + 5O2(g)  P4O10(s) What mass of phosphorus will be needed to produce 3.25 mol of P4O10? = 403 g P

  15. How to Solve Stoichiometric Problems - Streamlined • Convert given # into moles, if it isn’t already • Multiply by the mole ratio conversion factor • Convert from moles of substance into desired unit if necessary.

  16. + 2wheels + 1frame  1bike 45 g/wheel27 g/frame117 g/bike What mass of frames would be needed to ‘react’ completely with 3060 g wheels = 918 g frames

  17. 11.3 – Excess and Limiting Reactants The substance that is completely used up in a chemical rxn is called the limiting reactant. The substance that is NOT completely used up (and partially remains) is the excess reactant.

  18. Example: Copper reacts with sulfur to form copper(I) sulfide. What is the limiting reagent when 80.0 grams of Cu react with 25.0 g S? 2Cu + S  Cu2S 1. Convert given amounts into moles. 80.0 g Cu = 1.26 mol Cu = 0.75 mol S 24.0 g S 2. Multiply either amount by the mole ratio. 1 mol S  1.26 mol Cu = 0.630 mol Cu 2 mol Cu 3. The smaller number of mol is the limiting reactant. Cu is the limiting reagent

  19. You Try It! 2Fe + O2 + 2H2O  2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? 1. Calculate the amount of one reactant required to react with the other. 2. Compare the given amount to the required amount.

  20. actual yield percent yield =  100% theoretical yield 11.4 – Percent Yield The theoretical yield is the calculated amount of product that could be formed from given amounts of reactants; it is a the maximum amount. The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield; it is usually lower than the theoretical yield

  21. You Try It! What is the percent yield of CCl4 if 617 g is produced from the reaction of 312 g of CS2? CS2 + 3Cl2 CCl4 + S2Cl2

  22. Practice Quiz Quiz Practice “Studying” is a myth Learn from mistakes Best way to study is to re-do practice quiz Repeat Check answer key Some points can be recovered in 1:1 time with Tischer Use class time for work time and help

  23. Quarter Grades 50% Tests/Quizzes 35% Classwork/Homework 15% Labs Which of these matters the most?? Semester Grades 45% 1st Qtr 45% 2nd Qtr 10% Semester Test

  24. 11.5 – Molarity Molarity – unit of solution concentration 1 Liter = 1000 mL

  25. Calculate the molarity of these solutions. a) 0.55 mol NaOH dissolved in 1.0 L solution b) 4.0 grams of NaOH dissolved in 1.0 L solution c) 4.0 grams of NaOH dissolved in 250 mL solution

  26. How many moles of solute would be in the following solutions? a) 1.00 Liter of 2.2 M HNO3 b) 25.0 mL of 0.225 M HCl

  27. To make 1.0 L of a 1.0 M NaOH solution, 1.0 mol NaOH = 40.0 grams NaOH

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