1. KNOW HOW TO Percent error Percent yield Stoich Definitions T.Y LR STP Scenarios about pressure, volume, temperature Convert C to K and vise versa Lab Data stuff- stoich.

2. A student carried out a reaction between aqueous solution of lead II nitrate and potassium iodide. A double displacement reaction occurred when the solutions were poured together and filtered, a bright yellow precipitate of lead II iodide collected on the filter paper. The following data was collected during the experiment: Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3

3. 2. How much lead II nitrate used to start the reaction? Show work here: 41.26 g – 31.26 g = 10.00 g 3. How much potassium iodide used to start the reaction? Show work here: 46.26 g – 31.26 g = 15.00 g

4. Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 4. With the amount of lead II nitrate used at the start, calculate the theoretical yield of lead II iodide? 10.0 grams Pb(NO 3 ) 2 g of PbI 2 331.19 g of Pb(NO 3 ) 2 1 mole of Pb(NO 3 ) 2 461 g of PbI 2 1 mole of PbI 2 1 mole of Pb(NO 3 ) 2 1 mole of PbI 2 13.9 g of PbI 2

5. Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 5. With the amount of potassium iodide used at the start, calculate the theoretical yield of lead II iodide? 15.0 grams KI g of PbI 2 165.998 g of KI 1 mole of KI 461 g of PbI 2 1 mole of PbI 2 2 mole of KI 1 mole of PbI 2 20.8 g of PbI 2

6. Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 6. Which reactant is the limited reactant and its theoretical yield? Pb(NO 3 ) 2 and is theoretical yield is 13.9 grams

7. Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 7. Which reactant is in excess and by how much? 10.0 grams Pb(NO 3 ) 2 g of KI 331.19 g of Pb(NO 3 ) 2 1 mole of Pb(NO 3 ) 2 165.998 g of KI 1 mole of KI 1 mole of Pb(NO 3 ) 2 2 mole of KI 10.0 is used – 15.0 = 5.03 g of KI in excess 10.0 of KI used

8. Pb(NO 3 ) 2 + 2KI  PbI 2 + 2KNO 3 8. How much lead II iodide precipitate was actually collected from the lab?________________________________________ 12.7 g – 0.850g = 11.85 PbI 2 9. What is the percent yield? 11.85 grams 13.9 grams = 85.3%

9. CuCl 2 + Fe  FeCI 2 + Cu A single replacement reaction with Copper II chloride reacts with Iron metal (forms a +2 charge) to form two products,

10. 2. How much copper II chloride was used in the experiment? Show work here: 31.46 g – 56.46 g = 25.00 g 3. How much iron USED for the start the reaction? 18.51g – 10.41g = 8.10 grams USED in the reaction CuCl 2 + Fe  FeCI 2 + Cu

11. 4. With the amount of Copper II chloride used at the start, calculate the theoretical yield of copper metal? 25.0 grams CuCl 2 g of Cu 134.45 g of CuCl 2 1 mole of CuCl 2 63.55 g of Cu 1 mole of Cu 1 mole of CuCl 2 1 mole of Cu 11.8 g of Cu CuCl 2 + Fe  FeCI 2 + Cu

12. 5. With the amount of iron nails used at the start, calculate the theoretical yield of copper metal? 8.10 grams Fe g of Cu 55.85 g of Fe 1 mole of Fe 63.55 g of Cu 1 mole of Cu 1 mole of Fe 1 mole of Cu 9.22 g of Cu CuCl 2 + Fe  FeCI 2 + Cu

13. 6. Which reactant is the limited reactant and its theoretical yield? Fe and is theoretical yield is 9.22 grams CuCl 2 + Fe  FeCI 2 + Cu

14. 7. Which reactant is in excess and by how much? 8.10 g Fe g of CuCl2 55.85g Fe 1 mole of Fe 134.45 g of CuCl2 1 mole of CuCl2 1 mole of Fe 1 mole of CuCl 2 19.5 g of CuCl2 is used – 25.00 = 5.50 g of CuCl 2 in excess CuCl 2 + Fe  FeCI 2 + Cu

15. 9. What is the percent yield? 9.30 grams 9.22 grams = 100.87% WAY TOO MUCH- Contamination!!! CuCl 2 + Fe  FeCI 2 + Cu

17. What Gas Law? What Formula? 2. Helium gas in a 2.00-L cylinder is under 1.12 atm pressure. At 36.5ºC, that same gas sample has a pressure of 2.56 atm. What was the initial temperature (˚C) of the gas in the cylinder? Gay-Lussac’ Law P 1 = P 2 T1T1 T2T2

18. 2. Helium gas in a 2.00-L cylinder is under 1.12 atm pressure. At 36.5ºC, that same gas sample has a pressure of 2.56 atm. What was the initial temperature (˚C) of the gas in the cylinder? P 1 = P 2 T1T1 T2T2 1.12 atm = 2.56 atm X 309.5 K 346.6 2.56 = 135-273 = -138 ˚C

19. What Gas Law? What Formula? 4. Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00-L vessel at a pressure of 143 atm. Ideal PV = nRT

20. 4. Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00-L vessel at a pressure of 143 atm. PV = nRT (143 atm)(1.00L) = (2.49)(0.0821)(X K) 143 0.204429 = 699.5 - 273 = 427 ˚C

21. What Gas Law? What Formula? 6. Air trapped in a cylinder fitted with a piston occupies 145.7 mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? Boyles P 1 V 1 = P 2 V 2

22. 6. Air trapped in a cylinder fitted with a piston occupies 145.7 mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? P 1 V 1 = P 2 V 2 (1.08 atm)(145.7 ml) = (1.43)(X ml) 157.4 1.43 = 110 ml

23. What Gas Law? What Formula? 10. The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0ºC to 30.0ºC. What will be the resulting volume of this gas? Charles V 1 = V 2 T1T1 T2T2

24. 10. The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0ºC to 30.0ºC. What will be the resulting volume of this gas? 909 353 = 2.57 L V 1 = V 2 T1T1 T2T2 3.0 L 353 K303 K = X L

25. What Gas Law? What Formula? 12. A weather balloon contains 14.0 L of helium at a pressure of 95.5 kPa and a temperature of 12.0°C. If this had been stored in a 1.50-L cylinder at 21.0°C, what must the pressure in the cylinder have been? Combined P 1 V 1 = P 2 V 2 T1T1 T2T2

26. 12. A weather balloon contains 14.0 L of helium at a pressure of 95.5 kPa and a temperature of 12.0°C. If this had been stored in a 1.50-L cylinder at 21.0°C, what must the pressure in the cylinder have been? 393078 427.5 = 919 kPa P 1 V 1 = P 2 V 2 T1T1 T2T2 (14.0 L)(95.5 kPa) 285 K 294 K = (1.50 L)( X kPa)

27. I can… Apply the 3 gas laws to problems involving pressure, volume and temperature of a gas. Pressure v. Volume Temperature v. Volume Pressure v. Temperature

28. I can… Use the combined gas law equation with gas problems P 1 V 1 /T 1 = P 2 V 2 /T 2 Relate the amount of gas to its pressure, temperature and volume by using the ideal gas law PV = nRT