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Stoichiometry

Stoichiometry. What do I get when I have????. What is Stoichiometry?. a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions.

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Stoichiometry

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  1. Stoichiometry What do I get when I have????

  2. What is Stoichiometry? • a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions. • Stoichiometry can be used to calculate quantities such as the amount of products that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). In other words, • How much product do I get when I mix certain amounts of Reactants together, • Or, How much do I need to make what I want.

  3. Steps in Stoichmetry • Write the Balanced Chemical Equation. • Convert the grams given into moles. • Multiply by the mole ratio from the Balanced Chemical Equation. • Convert the new number of moles to grams.

  4. Why did I have to learn the Mole Equations? • The second step in Stoichiometry requires you to convert from grams to moles (or things to moles) • The fourth step in Stoichiometry requires you to convert from moles to grams (or moles to things)

  5. Railroad Track Stoichiometry

  6. The best way to learn is to do it • So here we go: You have 5 grams of Lead (II) Nitrate mixed with an excess of Sodium Chloride, how many grams of Lead (II) Chloride will you produce? • Step 1, write the balanced equation: • Pb(NO3)2+ 2 NaCl  PbCl2 + 2 NaNO3

  7. Step 2: Convert grams to moles • DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

  8. Step 3: Multiply by mole ratio • THIS IS THE ONLY STEP YOU USE THE COEFFIECENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  9. Step 4: Convert back to grams • DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

  10. And the Answer is: • Solve the problem

  11. One More time • You have 5 grams of Lead (II) Nitrate mixed with an excess of Sodium Chloride, how many grams of Sodium Nitrate will you produce? • Step 1, write the balanced equation: • Pb(NO3)2+ 2 NaCl  PbCl2 + 2 NaNO3

  12. Step 2: Convert grams to moles • DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

  13. Step 3: Multiply by mole ratio • THIS IS THE ONLY STEP YOU USE THE COEFFIECENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  14. Step 4: Multiply by New Molar Mass • DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

  15. And the Answer is:……… • Solve the problem

  16. Moles instead of grams??? • In case of this, you can skip either step 2 or step 4, YEAH!!!!!!!!!!!! • How many moles of Lead (II) Nitrate are needed to produce 6 grams of Sodium Nitrate? • Back to Step 1: • Pb(NO3)2+ 2 NaCl  PbCl2 + 2 NaNO3

  17. Starting in grams, so • We still need Step 2:

  18. Still have to do Step 3: • Moles to Moles:

  19. Back to the warm-up You work at a deli making sandwiches, your boss tells you that each sandwich requires 2 slices of bread, 3 slices of meat, and 1 slice of cheese. You have: -20 slices of bread -24 slices of meat -12 slices of cheese How many sandwiches can you make?

  20. Limiting Reagent The reactant that is completely consumed by the reaction The number of bicycles that can be assembled is limited by whichever part runs out first. In the inventory shown in this figure, wheels are that part.

  21. N2 + 3H2 2NH3 A molecular view of a Limiting reactant situation for the ammonia Synthesis. To make 4 molecules of NH3 requires 2 molecules of N2 and 6 molecules of H2. If we start with 4 molecules of N2 and 6 molecules of H2, H2 is the limiting reactant.

  22. Limiting Reagent Problems • Chemical Reactions will continue until one of the Reagents (Reactants) is used up. • Once one of the Reagents is used up, the reaction stops! • It doesn’t matter how much of the other one (Excess) is still left, the reactions STOPS.

  23. A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: • limiting reactant • amount of product C. Johannesson

  24. Limiting Reactants require TWO complete Stoichiometry Equations • One for each Reactant: • SAME FOUR STEPS, Two times……….. • If 5.00 grams of Lead (II) Nitrate is reacted with 5.00 grams of Sodium Chloride, what is the Limiting Reactant (Reagent), and how much of the solid product is produced? • Same STEP 1: • Pb(NO3)2+ 2 NaCl  PbCl2 (s) + 2 NaNO3

  25. It is a gram to gram problem, so… • You still have to do all the steps, but twice. • Once for each reactant: Pb(NO3)2 5.00 g Pb(NO3)2 PbCl2 331. 2 g Pb(NO3)2 PbCl2 Mole ratio from balanced equation

  26. Continuing: NaCl 5.00 g NaCl 11.9 PbCl2 PbCl2 58.443 g NaCl PbCl2 Mole ratio from balanced equation

  27. And……. • 5.00 g Lead (II) Nitrate makes 4.20 g of Lead (II) Chloride. • 5.00 g Sodium Chloride makes 11.9 grams of Lead (II) Chloride. • So, Lead (II) Nitrate is the Limiting Reagent because it makes the least Product. • Sodium Chloride is the Excess Reagent, it makes the most Product. • Only the Least Amount can be made, so only 4.20 g of Lead (II) Chloride is produced.

  28. YOU CAN ONLY MAKE THE … • LEAST AMOUNT OF PRODUCT. • A Car can only go as far as the Gasoline lets it go!!!!!!!!!!!

  29. Percent Yield • Stiochiometry gives us a perfect answer in a perfect world. • In the Real World not every atom reacts. • If you run your car out of gas, you can still smell gasoline in the gas tank, some it still left, but the car won’t run.

  30. % Yield = x 100 Actual Yield Theoretical Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. 3.10

  31. measured in lab calculated on paper B. Percent Yield C. Johannesson

  32. B. Percent Yield • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g C. Johannesson

  33. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: 45.8 g ? g actual: 46.3 g 1 mol K2CO3 138.204gK2CO3 45.8 g K2CO3 2 mol KCl 1 mol K2CO3 74.551 g KCl 1 mol KCl = 49.4 g KCl C. Johannesson

  34. 46.3 g 49.4 g B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield = 49.4 g KCl 45.8 g 49.4 g actual: 46.3 g  100 = 93.7% % Yield = C. Johannesson

  35. Example • In a chemical reaction between Lead (II) Nitrate and Sodium Chloride the Stoichiometry determined that 4.20 grams should be produced. When the Experiment was actually performed, 3.95 grams was produced. What is the Percent Yield? • % yield = 3.95 g actual X 100 = 94.05 % 4.20 g predicted

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