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Stoichiometry

Stoichiometry. Ch. 12 Page 352. What is “stoichiometry”?. A way of figuring out how much of a product can be made from a given amount of reactant Based on the Law of Conservation of Mass – in a chemical reaction, the mass of the reactants equals the mass of the products.

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Stoichiometry

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  1. Stoichiometry Ch. 12 Page 352

  2. What is “stoichiometry”? • A way of figuring out how much of a product can be made from a given amount of reactant • Based on the Law of Conservation of Mass – in a chemical reaction, the mass of the reactants equals the mass of the products.

  3. What do the coefficients in a chemical reaction represent? • The # of particles (atoms, formula units, molecules, ions) • The # of moles of each particle EX: 4Fe + 3O2 2Fe2O3 4 mol Fe 3 mol O2 2 mol Fe2O3 4 atoms Fe 3 molec. O2 2 formula units Fe2O3

  4. 4Fe + 3O2 2Fe2O3 • A balanced chemical equation ALSO tells us indirectly about the masses of the reactants and products (using the # moles of each substance and the molar masses). • How many grams of Fe are used in this reaction? • How many grams of O2? • How many grams of Fe2O3? (Let’s do these on the board.)

  5. 4Fe + 3O2 2Fe2O3 • Do our answers for the masses of the reactants and products in the equation above uphold the Law of Conservation of Mass? • YES! 4Fe + 3O2 2Fe2O3 (223.4 g + 96.0 g)  319.4 g 319.4 g = 319.4 g

  6. What is a “mole ratio”? • A balanced equation not only tells us the number of particles and number of moles. It also indicates the relationships among all substances involved. We use a mole ratio to show these relationships. • A mole ratio is a ratio between the # of moles of any two substances in a balanced chemical equation.

  7. Mole Ratio: the “Bridge” • MOLE RATIOS ARE THE KEY TO CALCULATIONS BASED UPON A CHEMICAL EQUATION! • From a balanced chemical equation, if you know the amount of one reactant, you can calculate the amount of any other reactant in the equation and the maximum amount of product you can obtain. • The “mole ratios” are the bridges we use to get from one kind of substance to another.

  8. 4Fe + 3O2 2Fe2O3 • How do you know how many mole ratios are available in an equation??? Just multiply the total number of substances (n) by the next lower number (n-1). • From this equation, there are 3 substances. Multiply 3 x 2 and you will get 6 available mole ratios. • Using the equation above, let’s write all the available mole ratios (on the board).

  9. Sec. 2: Using Stoichiometric Calculations • What are the tools needed for stoichiometric calculations? • A balanced chemical equation • Mole ratios • Mass-to-mole conversions

  10. Types of Stoichiometric Calculations • There are 3 types of stoichiometric calculations we will discuss: • mole-to-mole conversions • mole-to-mass conversions • mass-to-mass conversions For all of these conversions, we will use the following flow-chart:

  11. Grams of known (mass) Divide by Molar mass Moles of known Use “Mole Ratio” Moles of unknown Multiply by Molar mass Grams of unknown (mass)

  12. Steps in Stoichiometric Calculations • Write a balanced chemical equation. • Determine the moles of the known using the mass-mole conversion • Determine the moles of the unknown substance by using a mole ratio • Determine the mass of the unknown using a mole-mass conversion

  13. Complete the followingPractice Problems • Page 877: #5 – 10 • Page 379: #67 – 68 • Page 380: #72 -73

  14. Answers Pg. 877 • 0.600 mol HCl • 2.29 mol Cr2O3 • 924 g CO2 • 314 g H2SO4 • 1.31 g CO2 • 195.7 g aspirin

  15. Pg. 379 • (a.) K2CrO4 + Pb(NO3)  PbCrO4 + 2KNO3 (b.) 80.8 g PbCrO4 • (a.) N2H2 + H2O2 N2 + 2H2O (b.) 3.00 x 102 g N2H2

  16. Pg. 380 • (a.) Pb + PbO2 + 2H2SO4 2PbSO4 + 2H2O (b.) 73.2 g PbSO4 73. (a.) CO + 2H2 CH3OH (b.) 5.67 g H2

  17. 12.3: Limiting Reactants, pg. 364 • Rarely in nature are reactants in a chemical reaction present in the exact ratios specified by the balanced equation. • Generally, one or more reactants are in excess and the reaction proceeds until all of one reactant is used up.

  18. When a chemical reaction is carried out in the laboratory, the same principle applies. • Usually, one or more reactants are in excess, while one is limited. • The amount of product depends upon the reactant that is limited.

  19. The limiting reactant limits the extent of the reaction and, thereby, determines the amount of product. • A portion of the other reactants remains after the reaction stops. • These left-over reactants are called excess reactants.

  20. In section 2 of this chapter, we did calculations based on having the reactants present in the ratio described by the balanced chemical equation. • How can you calculate the amount of product formed when one reactant limits the amount of product and the other is in excess??

  21. Steps to determine the limiting reactant • create a balanced equation for the reaction 2. use stoichiometry to calculate how much product is produced by each reactant NOTE:  It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared. 3. The reactant that produces the lesser amount of product is the "limiting reactant."

  22. Example Limiting Reactant Calculation • A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which is the limiting reactant and how much excess reactant remains after the reaction has stopped?

  23. To determine the mass of the excess reactant • After determining the limiting reactant, do a mass of limiting reactantmass of excess reactant calculation (grams of limiting reactant  grams excess reactant) (THIS IS HOW MUCH OF THE EXCESS REACTANT ACTUALLY REACTS.) • Subtract that answer from the amount available in the reaction and you will have the mass of excess reactant left over.

  24. First, we need to create a balanced equation for the reaction: • 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

  25. Next we can use stoichiometry to calculate how much product is produced by each reactant.  NOTE:  It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared. The reactant that produces the lesser amount of product ,in this case is the oxygen, which is thus the "limiting reactant."

  26. Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen). We're not finished yet though. 

  27. 1.70 g is the amount of ammonia that reacted, not what is left over.  To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.

  28. Determining the Limiting Reactant • In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4).

  29. Additional Assessment Questions Topic 16 Question 1 Balance the following equation. How many moles of KClO3 are needed to produce 50 moles of O2?

  30. Answer

  31. 12.4: Percent Yield • In many calculations we have been practicing, we have been asked to calculate the amount of product that can be produced from a given amount of reactant. • The answer we obtained is called the theoretical yield: the max. amount of product that can be produced from a given amount of reactant.

  32. A chemical reaction rarely produces the theoretical yield of product. • When we conduct experiments, we determine the actual yield: the amount of product actually produced when the chemical reaction is carefully carried out in an experiment. • We measure efficiency by calculating percent yield with the following formula: % yield =__actual yield (from experiment)_ x 100 theoretical yield (from calculations)

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