1. 1 Chapter 12: Day 5 Ch12_stoic

2. 2 STOICHIOMETRY CALCULATIONS Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Molar mass given Molar mass Unknown

3. 3 PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much N 2 O and H 2 O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation NH 4 NO 3 ---> N 2 O + 2 H 2 O

4. 4 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 2 Convert mass reactant (454 g) --> moles STEP 3 Convert moles reactant (5.68 mol) --> moles product

5. 5 STEP 3 Convert moles reactant --> moles product Relate moles NH 4 NO 3 to moles product expected. 1 mol NH 4 NO 3 --> 2 mol H 2 O Express this relation as the STOICHIOMETRIC FACTOR.

6. 6 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O = 11.4 mol H 2 O produced STEP 3 Convert moles reactant (5.68 mol) --> moles product

7. 7 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 4 Convert moles product (11.4 mol) --> mass product Called the THEORETICAL YIELD ! ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

8. 8 0.21 mol AuCl 3 = 64 g x 1molAuCl 3 304 g AuCl 3 Mole ratio = 3Cl 2 2AuCl 3 = 0.32 mol Cl 2 X 71 g Cl 2 1mol Cl 2

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Gases in Equations The volume or amount of a gas in a chemical reaction can be calculated from the ideal gas law mole-mole factors from the balanced equation molar mass

10. 10 IDEAL GAS LAW n is proportional to V n is proportional to V (if T and P set) n is proportional to P (if V and T set) P V = n R T

11. 11 Mole ratio = 2SO 3 = VOLUME ratio = 12L 2SO 2 O 2 = VOLUME ratio = 6L 2H 2 SO 2 = VOLUME ratio = 22.4L O 2 4CO 2 = VOLUME ratio = 14L 2C 2 H 6 7O 2 = VOLUME ratio = 3.5ft 3 2C 2 H 6 In = 2+7 Out = 4+6 out > in

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12

13. 13 Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

14. 14 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) 1.1 g of H 2 O 2 is placed in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Strategy: Calculate moles of H 2 O 2 and then moles of O 2 and H 2 O. Finally, calc. P from n, R, T, and V. Strategy: Calculate moles of H 2 O 2 and then moles of O 2 and H 2 O. Finally, calc. P from n, R, T, and V.

15. 15 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Solution

16. 16 Gases and Stoichiometry Solution P of O 2 = 0.16 atm

17. 17 What is Pressue of H 2 O? Could calculate as above. OR recall Avogadro’s hypothesis. P  n at same T and V 2HO 2 = PRESSURE ratio X 0.16 atmO 2 1O 2 P of H 2 O = 0.32 atm 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g)

18. Basic Chemistry Copyright © 2011 Pearson Education, Inc. What volume, in L, of Cl 2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) STEP 1 Calculate the moles of given using molar mass or ideal gas law. 1 mol of Al = 26.98 g of Al 1 mol Al and 26.98 g Al 26.98 g Al 1 mol Al 1.50 g Al x 1 mol Al = 0.0556 mol of Al 26.98 g Al

19. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 19 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) STEP 2 Determine the moles of needed using a mole- RATIO 0.0556 mol Al x 3 mol Cl 2 = 0.0834 mol of Cl 2 2 mol Al

20. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 20 STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. To determine liters of gas, use the ideal gas law arranged to solve for V. T = 27 °C + 273 = 300. K V = nRT = (0.0834 mol Cl 2 )(0.0821 L atm/mol K)(300. K) P 1.20 atm = 1.71 L of Cl 2

21. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 21 What volume (L) of O 2 at 24 °C and 0.950 atm is needed to react with 28.0 g of NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

22. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 22 STEP 1 Calculate the moles of given using molar mass or ideal gas law. 1 mol of NH 3 = 17.03 g of NH 3 1 mol NH 3 and 17.03 g NH 3 17.03 g NH 3 Al 1 mol NH 3 28.0 g NH 3 x 1 mol NH 3 = 1.64 mol of NH 3 17.03 g NH 3 Solution

23. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 23 STEP 2 Determine the moles of needed using a mole-mole factor. 5 mol of O 2 = 4 mol of NH 3 4 mol NH 3 and 5 mol O 2 5 mol O 2 4 mol NH 3 1.64 mol NH 3 x 5 mol O 2 = 2.05 mol of O 2 4 mol NH 3 Solution (continued)

24. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 24 STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. To determine liters of gas, use the ideal gas law arranged to solve for V. T = 24 °C + 273 = 297 K Place the moles of O 2 in the ideal gas law. V = nRT =(2.05 mol)(0.0821 L atm/mol K)(297 K) P 0.950 atm = 52.6 L of O 2 Solution (continued)

25. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 25 What mass of Fe will react with 5.50 L of O 2 at STP? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 1) 13.7 g of Fe 2) 18.3 g of Fe 3) 419 g of Fe Learning Check

26. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 26 STEP 1 Calculate the moles of given using molar mass or ideal gas law. Use molar volume at STP to calculate moles of O 2. 5.50 L O 2 x 1 mol O 2 = 0.246 mol of O 2 22.4 L O 2 STEP 2 Determine the moles of needed using a mole- mole factor. 4 mol of Fe = 3 mol of O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe 0.246 mol O 2 x 4 mol Fe = 0.328 mol of Fe 3 mol O 2 Solution

27. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 27 STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. 1 mol of Fe = 55.85 g of Fe 1 mol Fe and 55.85 g Fe 55.85 g Fe 1 mol Fe 0.328 mol Fe x 55.85 g Fe = 18.3 g of Fe 1 mol Fe Placing all three steps in one setup gives (STEP 1) (STEP 2) (STEP 3) 5.50 L O 2 x 1 mol O 2 x 4 mol Fe x 55.85 g Fe = 18.3 g of Fe 22.4 L O 2 3 mol O 2 1 mol Fe Solution (continued)