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Chemical Quantities

Chemical Quantities. Chapter 9. Chemical Stoichiometry. Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. Stoichiometry is used to determine how much stomach acid an antacid tablet can neutralize. Chemical Equations.

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Chemical Quantities

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  1. Chemical Quantities • Chapter 9

  2. Chemical Stoichiometry • Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. • Stoichiometry is used to determine how much stomach acid an antacid tablet can neutralize.

  3. Chemical Equations A balanced chemical equation is like a recipe. One needs to know what the ingredients are and what relative amounts of ingredients are needed for both recipes and chemical equations.

  4. Chemical Equation • A representation of a chemical reaction: • C2H5OH(l) + __O2(g)__CO2(g) + __H2O(g) • reactants products • Is this equation balanced? • C2H5OH(l) + 3O2(g)2CO2(g) + 3H2O(g)

  5. Chemical Equation • C2H5OH (l)+3O2(g)2CO2(g)+3H2O(g) • Microscopic: • 1 molecule of ethanol reacts with3 molecules of oxygento produce 2 molecules of carbon dioxide and3 molecules of water. • Macroscopic: • 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

  6. Chemical Equation • C2H5OH (l)+3O2(g)2CO2(g)+3H2O(g) • Microscopic: • 1 dozen molecules of ethanol reacts with3 dozen molecules of oxygento produce 2 dozen molecules of carbon dioxide and3 dozen molecules of water. • Macroscopic: • 6.02 x 1023 molecules of ethanol reacts with 3(6.02 x 1023)molecules of oxygen to produce 2(6.02 x 1023)molecules of carbon dioxide and 3(6.02 x 1023 ) molecules of water.

  7. Chemical Equation • C2H5OH (l)+3O2(g)2CO2(g)+3H2O(g) • Macroscopic: • 46.0 g of ethanol reacts with 96.0g of oxygen to produce 88.0 g of carbon dioxide and 54.0 g of water. • 142.0 g of reactants = 142.0 g of products. • Atoms and mass are conserved in a chemical reaction, but moles and molecules are not!!!

  8. Moles & Molecules • __C3H8(g) + __O2(g) ---> __CO2(g) + __HOH(g) • C3H8(g) + 5O2(g) ---> 3CO2(g) + 4HOH(g) • How many moles and molecules of each substance are there?

  9. Mole Ratio • Mole ratio -- a conversion factor based upon a balanced equation and used to determine relative amounts of reactants and products. • Mole ratios can exist between a reactant and a product, between two reactants, or between two products. • C3H8(g) + 5O2(g) ---> 3CO2(g) + 4HOH(g)

  10. Types of Stoichiometry Problems • Mole to Mole • Gram to Mole • Mole to Gram • Gram to Molecules • Molecules to Gram

  11. Calculating Masses of Reactants and Products • 1. Balance the equation. • 2. Convert mass to moles. • 3. Set up mole ratios. • 4. Use mole ratios to calculate moles of desired reactant or product. • 5. Convert moles to grams, if necessary.

  12. Mole To Mole Problems • What number of moles of oxygen would be used in burning 5.8 moles of propane, C3H8? • C3H8(g) + 5O2(g) ---> 3CO2(g) + 4HOH(g) • (5.8 mol C3H8)(5 mol O2/1 mol C3H8) = • 29 mol O2

  13. Gram to Mole & Gram to Gram • __Al(s) + __I2(s) ---> __AlI3(s) • 2Al(s) + 3I2(s) ---> 2AlI3(s) • How many moles and how many grams of aluminum iodide can be produce from 35.0 g of aluminum?

  14. Gram to Mole & Gram to Gram • 2Al(s) + 3I2(s) ---> 2AlI3(s) • (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al) = 1.30 mol AlI3 • (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al)(407.68 g/1 mol) = 529 g AlI3

  15. Grams to Molecules • __LiOH(s) + __CO2(g) ---> __Li2CO3(s) + __HOH(l) • 2LiOH(s) + CO2(g) ---> Li2CO3(s) + HOH(l) • How many molecules of water would be formed from 1.00 x 103 g of LiOH? • (1.00 x 103 g LiOH)(1 mol/23.95 g)(1 mol HOH/ • 2 mol LiOH)(6.02 x 1023 molecules/1 mol) • = 1.26 x 1025 molecules HOH

  16. Stoichiometric Quantities • Stoichiometric Quantities -- quantities of reactants mixed in exactly the amounts that result in their all being used up at the same time. • How often do you think this occurs in reality? • Almost never!!!!

  17. Limiting Reactant The limiting reactant is the reactant that is consumed first,limiting the amounts of products formed. Almost all stoichiometric situations are of the limiting reactant type. The reactants that are left over and unreacted are said to be in excess.

  18. Figure 9.1: A mixture of 5CH4 and 3H20 molecules undergoes the reaction CH4(g) + H20(g) ---> 3H2 + CO(g)

  19. Double Cheeseburger Problem • At the local “Burger Barn” a worker finds the following inventory: • 22 hamburger patties • 15 hamburger buns • 7 slices of onion • 18 slices of cheese • How many double cheeseburgers with onion and cheese can be made to sell?

  20. Double Cheeseburger Problem • 2 h.b. patties + 1 h.b. bun + 2 slices cheese + • 1 slice onion ---> 1 double cheeseburger • What is the limiting reactant? • Onion • How many double cheeseburgers with onion and cheese can be made to sell? • 7 double cheeseburgers

  21. Double Cheeseburger Problem • 2 h.b. patties + 1 h.b. bun + 2 slices cheese + • 1 slice onion ---> 1 double cheeseburger • What materials are in excess and by how much? • 8 hamburger patties • 8 hamburger buns • 4 slices cheese

  22. Solving a Stoichiometry Problem • 1. Balance the equation. • 2. Convert masses to moles. • 3. Determine which reactant is limiting. • 4. Use moles of limiting reactant and mole ratios to find moles of desired product. • 5. Convert from moles to grams.

  23. Limiting Reactant Problem • If 56.0 g of Li reacts with 56.0 g of N2, how many grams of Li3N can be produced? • __Li(s) + __N2(g) ---> __Li3N(s) • 6 Li(s) + N2(g) ---> 2 Li3N(s) • (56.0 g Li) (1 mol/6.94g)(1 mol N2/6 mol Li) (28.0 g/1 mol) = 37.7 g N2 • Since there were 56.0 g of N2 and only 37.7 g used, N2 is the excess and Li is the Limiting Reactant.

  24. Limiting Reactant Problem • 6 Li(s) + N2(g) ---> 2 Li3N(s) • (56.0 g Li)(1 mol/6.94g)(2 mol Li3N/6 mol Li) (34.8 g/1 mol) = 93.6 g Li3N • How many grams of nitrogen are left? • 56.0g N2 given - 37.7 g used = 18.3 g excessN2

  25. Double Cheeseburger Yield • At the local “Burger Barn” a worker finds the following inventory: • 22 hamburger patties • 15 hamburger buns • 7 slices of onion • 18 slices of cheese • We found that seven double cheeseburgers could be made from these ingredients.

  26. Double Cheeseburger Yield • If a worker eats one slice of onion, how many double cheeseburgers can actually be made? • 6 double cheeseburgers • The number of cheeseburgers that could have been made (7) is the theoretical yield. • The number of cheeseburgers that actuallywere made (6) is the actual yield.

  27. Double Cheeseburger Yield • % Yield = 85.7 %

  28. % Yield • Values calculated using stoichiometry are always theoretical yields! • Values determined experimentally in the laboratory are actual yields!

  29. Limiting Reactant & % Yield • If 68.5 kg of CO(g) is reacted with 8.60 kg of H2(g), what is the theoretical yield of methanol that can be produced? • __H2(g) + __CO(g) ---> __CH3OH(l) • 2 H2(g) + CO(g) ---> CH3OH(l) • (68.5 kg CO)(1 mol/28.0 g)(2 mol H2/1 mol CO) • (2.02 g/1mol) = 9.88 kg H2

  30. Limiting Reactant & % Yield • 2 H2(g) + CO(g) ---> CH3OH(l) • Since only 8.60 kg of H2 were provided, the H2 is the limiting reactant, and the CO is in excess. • (8.60 kg H2)(1000 g/1 kg)(1 mol/2.02 g)(1 mol CH3OH/2 mol H2)(32.0 g/1 mol) = 6.85 x 104 g CH3OH

  31. Limiting Reactant & % Yield • 2 H2(g) + CO(g) ---> CH3OH(l) • If in the laboratory only 3.57 x 104 g of CH3OH is produced, what is the % yield? % Yield = 52.1 %

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