1. Entry Task: March 13-14 th 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. 25.0 g NaBr 14 g NaCl 103 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 1 mole NaBr 1 mol NaCl 20 g KCl 15.6 g NaCl 74.5g KCl 1 mole KCl58.4 g NaCl 1 mole NaCl 1 mole KCl 1 mole NaCl

7. Which reactant is Limited and which reactant is Excess? Limited = NaBr Excess = KCl ___NaBr + ___KCl  ___KBr + ___NaCl 25.0 g NaBr 14 g NaCl 103 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 1 mole NaBr 1 mol NaCl 20 g KCl 15.6 g NaCl 74.5g KCl 1 mole KCl58.4 g NaCl 1 mole NaCl 1 mole 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. Start with the limited reactant!! Used In reaction 20 grams KCl minus 18 g KCl = Given Used In reaction 2 g EXCESS KCl ___NaBr + ___KCl  ___KBr + ___NaCl 25.0 g NaBr 18 g KCl 103 g NaBr 1 mole NaBr 74.5g KCl 1 mole KCl 1 mole N aBr 1 mole KCl

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

10. ___NaBr + ___KCl  ___KBr + ___NaCl 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. 50.0 g NaBr 28.3 g NaCl 103 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 1 mole NaBr 1 mol NaCl 25.0 g KCl 19.6 g NaCl 74.5g KCl 1 mole KCl58.4 g NaCl 1 mole NaCl 1 mole KCl 1 mole NaCl

11. Which reactant is Limited and which reactant is Excess? Limited = KCl Excess = NaBr ___NaBr + ___KCl  ___KBr + ___NaCl 50.0 g NaBr 28.3 g NaCl 103 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 1 mole NaBr 1 mol NaCl 25.0 g KCl 19.6 g NaCl 74.5g KCl 1 mole KCl58.4 g NaCl 1 mole NaCl 1 mole 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. Start with the limited reactant!! Used In reaction 50.0 grams NaBr minus 34.6 g NaBr = Given Used In reaction 15.4 g EXCESS NaBr ___NaBr + ___KCl  ___KBr + ___NaCl 25.0 g KCl 34.6 g NaBr 74.5g KCl 1 mole KCl103 g NaBr 1 mole NaBr 1 mole KCl 1 mole NaBr

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

14. 25 grams of lead II iodide reacts with potassium nitrate to create lead II nitrate and potassium iodide. ___Pb I 2 + ___KNO 3  ___Pb(NO 3 ) 2 + ___K I 22

15. ___Pb I 2 + _2_KNO 3  ___Pb(NO 3 ) 2 + _2_K I 1. How much potassium nitrate is needed if 25 grams of lead II iodide were used in the reaction? 25.0 g Pb I 2 11.0 g KNO 3 461 g Pb I 2 1 mole Pb I 2 101 g KNO 3 1 mol KNO 3 1 mole Pb I 2 2 mol KNO 3 I needed to find out how much of the OTHER reactant I need to complete this reaction. Compare reactant to reactant !!

16. ___Pb I 2 + _2_KNO 3  ___Pb(NO 3 ) 2 + _2_K I 2. How much potassium iodide was produced from the reaction with 25 grams of lead II iodide used? 25.0 g Pb I 2 18.0 g K I 461 g Pb I 2 1 mole Pb I 2 166 g K I 1 mol K I 1 mole Pb I 2 2 mol K I Now I’m finding out how much product- KI is produced with my 25 grams of PbI 2. Compare reactant to product !!

17. ___Pb I 2 + _2_KNO 3  ___Pb(NO 3 ) 2 + _2_K I 3. Using the amount of potassium nitrate, how much potassium iodide would be produced? 11.0 g KNO 3 18.0 g K I 101 g KNO 3 1 mole KNO 3 166 g K I 1 mol K I 2 mole KNO 3 2 mol K I NOW I’m using the other reactant KNO 3 amount that we figured out from the ratio with PbI 2. Compare reactant to reactant !!

18. ___Pb I 2 + _2_KNO 3  ___Pb(NO 3 ) 2 + _2_K I 4. What can you deduce from your answers from 2 and 3? 11.0 g KNO 3 18.0 g K I 101 g KNO 3 1 mole KNO 3 166 g K I 1 mol K I 2 mole KNO 3 2 mol K I 25.0 g Pb I 2 18.0 g K I 461 g Pb I 2 1 mole Pb I 2 166 g K I 1 mol K I 1 mole Pb I 2 2 mol K I By reacting 25 g of Pb I 2 and 11 g of KNO 3, you will produce 18 g of K I NO WASTE!!.

19. ___Pb I 2 + _2_KNO 3  ___Pb(NO 3 ) 2 + _2_K I 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 100 g KNO 3 164 g KI 101 g KNO 3 1 mol KNO 3 166 g KI 1 mol KI 2 mol KNO 3 2 mol KI 100 g KNO 3 164 g Pb(NO 3 ) 2 101 g KNO 3 1 mol KNO 3 331.12g Pb(NO 3 ) 2 1 mole Pb(NO 3 ) 2 2 mol KNO 3 1 mole Pb(NO 3 ) 2 LETS CHECK OUR WORK!!!

20. ___Pb I 2 + _2_KNO 3  ___Pb(NO 3 ) 2 + _2_K I 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 164 g KI 99.88 g KNO 3 166 g KI 1 mol KI 101.1 g KNO 3 1 mol KNO 3 2 mol KI 2 mol KNO 3 164 g Pb(NO 3 ) 2 100g KNO 3 331.12g Pb(NO 3 ) 2 1 mol Pb(NO 3 ) 2 101 g KNO 3 1 mole KNO 3 1 mol Pb(NO 3 ) 2 2 mole KNO 3