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Entry Task: March 13-14 th Block 2

Entry Task: March 13-14 th Block 2. Agenda:. Discuss Limited and % yield ws Self Check on Limited reactant & % yields. I can…. Use the sequence of steps used in solving stoichiometric problems Identify the limiting reactant in a chemical equation

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Entry Task: March 13-14 th Block 2

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  1. Entry Task: March 13-14th Block 2

  2. Agenda: • Discuss Limited and % yield ws • Self Check on Limited reactant & % yields

  3. I can… • Use the sequence of steps used in solving stoichiometric problems • Identify the limiting reactant in a chemical equation • Identify the excess reactant and calculate the amount remaining after the reaction is complete. • Calculate the mass of a product when the amounts of more that one reactant are given • Calculate the theoretical yield of a chemical reaction from data. • Determine the percent yield for a chemical reaction.

  4. Limited reactants % yields practice

  5. A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl ___KBr + ___NaCl

  6. A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl  ___KBr + ___NaCl 25.0 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 14 gNaCl 103 g NaBr 1 mole NaBr 1 mol NaCl 20 g KCl 1 mole KCl 1 mole NaCl 58.4 g NaCl 15.6 g NaCl 1 mole KCl 74.5g KCl 1 mole NaCl

  7. Which reactant is Limited and which reactant is Excess? ___NaBr + ___KCl  ___KBr + ___NaCl Limited= NaBr 25.0 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 14 gNaCl 103 g NaBr 1 mole NaBr 1 mol NaCl Excess = KCl 20 g KCl 1 mole KCl 1 mole NaCl 58.4 g NaCl 15.6 g NaCl 1 mole KCl 74.5g KCl 1 mole NaCl

  8. A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl  ___KBr + ___NaCl Start with the limited reactant!! Used In reaction 25.0 g NaBr 1 moleNaBr 74.5g KCl 1 mole KCl 18 g KCl 1 mole NaBr 103 g NaBr 1 mole KCl Given Used In reaction 20 grams KClminus 18 g KCl = 2 g EXCESS KCl

  9. If 12 grams of sodium chloride were produced after the reaction, what is the % yield? • Actual yield (from experiment) • Percent Yield= • X 100 • Theoretical yield (from stoich calculation) • 12 g (from experiment) • Percent Yield= • X 100 • 14 g (from stoich calculation-limit reactant) • 86%

  10. With the same reaction, what if 50.0 grams of sodium bromide reacted with 75.0 grams of potassium chloride. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl  ___KBr + ___NaCl 50.0 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 28.3 gNaCl 103 g NaBr 1 mole NaBr 1 mol NaCl 25.0 g KCl 1 mole KCl 1 mole NaCl 58.4 g NaCl 19.6 g NaCl 1 mole KCl 74.5g KCl 1 mole NaCl

  11. Which reactant is Limited and which reactant is Excess? ___NaBr + ___KCl  ___KBr + ___NaCl Excess = NaBr 50.0 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 28.3 gNaCl 103 g NaBr 1 mole NaBr 1 mol NaCl Limited= KCl 25.0 g KCl 1 mole KCl 1 mole NaCl 58.4 g NaCl 19.6 g NaCl 1 mole KCl 74.5g KCl 1 mole NaCl

  12. With the same reaction, what if 50.0 grams of sodium bromide reacted with 25.0 grams of potassium chloride. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl  ___KBr + ___NaCl Start with the limited reactant!! Used In reaction 25.0 g KCl 1 moleKCl 103 g NaBr 1 mole NaBr 34.6 g NaBr 1 mole KCl 74.5g KCl 1 mole NaBr Given Used In reaction 50.0 grams NaBrminus 34.6 g NaBr = 15.4 g EXCESS NaBr

  13. If 14 grams of sodium chloride were produced after the reaction, what is the % yield? • Actual yield (from experiment) • Percent Yield= • X 100 • Theoretical yield (from stoich calculation) • 14 g (from experiment) • Percent Yield= • X 100 • 19.6 g (from stoich calculation-limit reactant) • 71%

  14. 25 grams of lead II iodide reacts with potassium nitrate to create lead II nitrate and potassium iodide. ___PbI2 + ___KNO3 ___Pb(NO3)2 + ___KI 2 2

  15. 1. How much potassium nitrate is needed if 25 grams of lead II iodide were used in the reaction? ___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI 25.0 g PbI2 1 mole PbI2 101 g KNO3 2 mol KNO3 11.0 gKNO3 461 g PbI2 1 mole PbI2 1 mol KNO3 I needed to find out how much of the OTHER reactant I need to complete this reaction. Compare reactant to reactant!!

  16. 2. How much potassium iodide was produced from the reaction with 25 grams of lead II iodide used? ___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI 25.0 g PbI2 1 mole PbI2 166 g KI 2 mol KI 18.0 gKI 461 g PbI2 1 mole PbI2 1 mol KI Now I’m finding out how much product- KI is produced with my 25 grams of PbI2. Compare reactant to product!!

  17. 3. Using the amount of potassium nitrate, how much potassium iodide would be produced? ___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI 11.0 g KNO3 1 mole KNO3 166 g KI 2 mol KI 18.0 gKI 101 g KNO3 2 mole KNO3 1 mol KI NOW I’m using the other reactant KNO3 amount that we figured out from the ratio with PbI2. Compare reactant to reactant!!

  18. 4. What can you deduce from your answers from 2 and 3? ___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI 25.0 g PbI2 1 mole PbI2 166 g KI 2 mol KI 18.0 gKI 461 g PbI2 1 mole PbI2 1 mol KI 11.0 g KNO3 1 mole KNO3 166 g KI 2 mol KI 18.0 gKI 101 g KNO3 2 mole KNO3 1 mol KI By reacting 25 g of PbI2 and 11 g of KNO3, you will produce 18 g of KI NO WASTE!!.

  19. 5. What if I need 100 grams of potassium nitrate to be produced? How much of each reactant would be needed to complete this reaction? Work backwards ___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI 100 g KNO3 166g KI 2 molKI 1 molKNO3 164g KI 101 g KNO3 2 molKNO3 1 molKI 331.12g Pb(NO3)2 1 molKNO3 1 mole Pb(NO3)2 100 g KNO3 164 g Pb(NO3)2 2 molKNO3 101 g KNO3 1 mole Pb(NO3)2 LETS CHECK OUR WORK!!!

  20. 5. What if I need 100 grams of potassium nitrate to be produced? How much of each reactant would be needed to complete this reaction? Work backwards ___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI 164 g KI 101.1g KNO3 2 mol KNO3 1 molKI 99.88g KNO3 166 g KI 2 molKI 1 molKNO3 1 molPb(NO3)2 2 mole KNO3 164 g Pb(NO3)2 101 g KNO3 100g KNO3 1 molPb(NO3)2 331.12g Pb(NO3)2 1 mole KNO3

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