2. Entry Task: March 11 th Monday Entry task question: 2AgCl + Mg  MgCl 2 + 2Ag Question: How many grams of silver can be produced if 500.0 grams of magnesium reacts with excess AgCl? You have ~10 minutes

3. Agenda: Sign off and Discuss Stoichiometry #3 ws Notes on Limited Reactants

5. Copper II sulfate reacts with potassium phosphate to create copper II phosphate and potassium sulfate ___CuSO 4 + ___K 3 PO 4  ___K 2 SO 4 + ___Cu 3 ( PO 4 ) 2 3 32

6. How many grams of copper II sulfate is needed to produce 45.0g of potassium sulfate? 45.0 grams K 2 SO 4 g of CuSO 4 174.257 g of K 2 SO 4 1 mole of K 2 SO 4 159.54 g of CuSO 4 1 mole of CuSO 4 3 mole of K 2 SO 4 3 mole of CuSO 4 21537.9 522.77 41.2 g of CuSO 4 ___CuSO 4 + ___K 3 PO 4  ___K 2 SO 4 + ___Cu 3 ( PO 4 ) 2 3 32

7. If 74.9 grams of potassium sulfate is produced from this reaction, how much copper II phosphate can be produced? 74.9 grams K 2 SO 4 g of Cu 3 (PO 4 ) 2 174.287 g of K 2 SO 4 1 mole of K 2 SO 4 380.578g of Cu 3 (PO 4 ) 2 1 mole of Cu 3 (PO 4 ) 2 3 mole of K 2 SO 4 1 mole of Cu 3 (PO 4 ) 2 28505.29 522.861 54.5 of Cu 3 (PO 4 ) 2 ___CuSO 4 + ___K 3 PO 4  ___K 2 SO 4 + ___Cu 3 ( PO 4 ) 2 3 32

8. How many grams of potassium phosphate are needed to react with 75.0 grams of copper II sulfate to complete the reaction? 75.0 grams CuSO 4 g of K 3 PO 4 159.54 g of CuSO 4 1 mole of CuSO 4 211.97g of K 3 PO 4 1 mole of K 3 PO 4 3 mole of CuSO 4 2 mole of K 3 PO 4 31795.5 478.62 66.5 g of K 3 PO 4 ___CuSO 4 + ___K 3 PO 4  ___K 2 SO 4 + ___Cu 3 ( PO 4 ) 2 3 32

9. If there are 2.09 grams of potassium phosphate produced in this reaction, how much copper II sulfate is required? 2.09 grams K 3 PO 4 g of CuSO 4 211.97 g of K 3 PO 4 1 mole of K 3 PO 4 159.54g of CuSO 4 1 mole of CuSO 4 2 mole of K 3 PO 4 3 mole of CuSO 4 1000.32 423.94 2.36 g of CuSO 4 ___CuSO 4 + ___K 3 PO 4  ___K 2 SO 4 + ___Cu 3 ( PO 4 ) 2 3 32

10. How many grams of potassium phosphate will be needed to produce 64.5 grams of copper II phosphate? 64.5 grams Cu 3 (PO 4 ) 2 g of K 3 PO 4 380.58 g of Cu 3 (PO 4 ) 2 1 mole of Cu 3 (PO 4 ) 2 211.97g of K 3 PO 4 1 mole of K 3 PO 4 1 mole of Cu 3 (PO 4 ) 2 2 mole of K 3 PO 4 27344.13 380.58 71.8 g of K 3 PO 4 ___CuSO 4 + ___K 3 PO 4  ___K 2 SO 4 + ___Cu 3 ( PO 4 ) 2 3 32

11. If there are 1.25 grams of copper II phosphate produced in a reaction, how much potassium phosphate is required? 1.25 grams Cu 3 (PO 4 ) 2 g of K 3 PO 4 380.58 g of Cu 3 (PO 4 ) 2 1 mole of Cu 3 (PO 4 ) 2 211.97g of K 3 PO 4 1 mole of K 3 PO 4 1 mole of Cu 3 (PO 4 ) 2 2 mole of K 3 PO 4 529.25 380.58 1.39 g of K 3 PO 4 ___CuSO 4 + ___K 3 PO 4  ___K 2 SO 4 + ___Cu 3 ( PO 4 ) 2 3 32

12. I can 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

14. Limiting Reactant There are 2 types of reactants in a chemical reaction There are 2 types of reactants in a chemical reaction After a reaction, the reactant is said to be in excess (there is too much). After a reaction, the reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once this reactant runs out, the reaction s. This is called the limiting reactant. The other reactant limits how much product we get. Once this reactant runs out, the reaction s. This is called the limiting reactant.

15. Limiting Reactant 1.To find out which reactant in the chemical reaction is the limiting reactant, we have to try each reactant involved in the reaction. 2.We have to calculate how much of product we can get from each reactant to determine which is the limiting one. 3.The lower amount of product is the correct reactant (limited reactant). 4.When you pick a product for your calculations, use the same product in your comparison.

16. 2 Al + 3 Cl 2  2 AlCl 3 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 10.0 g Al 49.4 g AlCl 3 27.0 g Al 1 mole Al133.5 g AlCl 3 1 mol AlCl 3 2 mole Al 2 mol AlCl 3 35.0g Cl 2 43.5 g AlCl 3 71.0 g Cl 2 1 mole Cl 2 133.5 g AlCl 3 1 mole AlCl 3 3 mole Cl 2 2 mole AlCl 3

17. Which reactant is Limited and which reactant is Excess? Limited = Cl 2 Excess = Al 10.0 g Al 49.4 g AlCl 3 27.0 g Al 1 mole Al133.5 g AlCl 3 1 mol AlCl 3 2 mole Al 2 mol AlCl 3 35.0g Cl 2 43.5 g AlCl 3 71.0 g Cl 2 1 mole Cl 2 133.5 g AlCl 3 1 mole AlCl 3 3 mole Cl 2 2 mole AlCl 3 2 Al + 3 Cl 2  2 AlCl 3

18. 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? Start with the limited reactant!! Used In reaction 10.0 grams Al minus 8.9 g Al = Given Used In reaction 1.1 g EXCESS 2 Al + 3 Cl 2  2 AlCl 3 35.0g Cl 2 8.9 g Al 71.0 g Cl 2 1 mole Cl 2 26.98 g Al 1 mole Al 3 mole Cl 2 2 mole Al

19. 2 K + I 2  2 K I 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made AND how much excess. 15.0 g K 63.68 g KI 39.098 g K 1 mole K165.99 g KI 1 mol KI 2 mole K 2 mol KI 15.0 g I 2 19.6 g K I 253.8 g I 2 1 mol I 2 165.99 g KI 1 mol KI 1 mol I 2 2 mol KI

20. Which reactant is Limited and which reactant is Excess? Limited = l 2 Excess = K 2 K + I 2  2 K I 15.0 g K 63.68 g KI 39.098 g K 1 mole K165.99 g KI 1 mol KI 2 mole K 2 mol KI 15.0 g I 2 19.6 g K I 253.8 g I 2 1 mol I 2 165.99 g KI 1 mol KI 1 mol I 2 2 mol KI

21. 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made AND how much excess. Start with the limited reactant!! Used In reaction 15.0 grams K minus 4.62 g K = Given Used In reaction 10.38 g EXCESS 15.0 g I 2 4.62 g K 253.8 g I 2 1 mol I 2 39.1 g K 1 mol K1 mol I 2 2 mol K 2 K + I 2  2 K I

22. Limiting Reactant: Recap 1.You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2.Convert ALL of the reactants to the SAME product (pick any product you choose.) 3.The lowest amount of product involved the reactant which is the LIMITING REACTANT. 4.The other reactant(s) are in EXCESS. 5.Start with limited reactant and compare it with the other reactants in a gram to gram Stoichiometry conversation. 6.Then find the amount of excess, subtract the amount used from the given amount.