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Chapter 12: Day 5

Chapter 12: Day 5. Ch12_stoic. Mass product. Mass reactant. Moles reactant. Moles product. STOICHIOMETRY CALCULATIONS. Molar mass Unknown. Molar mass given. Stoichiometric factor.

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Chapter 12: Day 5

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  1. Chapter 12: Day 5 Ch12_stoic

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

  3. PROBLEM: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation NH4NO3 ---> N2O + 2 H2O

  4. 454 g of NH4NO3 --> N2O + 2 H2O STEP 2 Convert mass reactant (454 g) --> moles STEP 3 Convert moles reactant (5.68 mol) --> moles product

  5. STEP 3 Convert moles reactant --> moles product Relate moles NH4NO3 to moles product expected. 1 mol NH4NO3 --> 2 mol H2O Express this relation as the STOICHIOMETRIC FACTOR.

  6. 454 g of NH4NO3 --> N2O + 2 H2O STEP 3 Convert moles reactant (5.68 mol) --> moles product = 11.4 mol H2O produced

  7. 454 g of NH4NO3 --> N2O + 2 H2O STEP 4 Convert moles product (11.4 mol) --> mass product Called the THEORETICAL YIELD ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

  8. 0.21 molAuCl3 = 64 g x 1molAuCl3304 g AuCl3 Mole ratio = 3Cl2 2AuCl3 X 71 g Cl21mol Cl2 = 0.32 mol Cl2

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

  11. Mole ratio = 2SO3 = VOLUME ratio = 12L 2SO2 O2 = VOLUME ratio = 6L 2H2 SO2 = VOLUME ratio = 22.4L O2 4CO2 = VOLUME ratio = 14L 2C2H6 7O2 = VOLUME ratio = 3.5ft3 2C2H6 In = 2+7 Out = 4+6 out > in

  12. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

  13. 2 H2O2(liq) ---> 2 H2O(g) + O2(g) 1.1 g of H2O2 is placed in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Strategy: Calculate moles of H2O2 and then moles of O2 and H2O. Finally, calc. P from n, R, T, and V.

  14. 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Solution

  15. Gases and Stoichiometry Solution P of O2 = 0.16 atm

  16. 2 H2O2(liq) ---> 2 H2O(g) + O2(g) What is Pressue of H2O? Could calculate as above. OR recall Avogadro’s hypothesis. P n at same T and V • 2HO2= PRESSURE ratio X 0.16 atmO2 1O2 P of H2O = 0.32 atm

  17. What volume, in L, of Cl2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of aluminum? 2Al(s) + 3Cl2(g) 2AlCl3(s) STEP 1Calculate 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

  18. 2Al(s) + 3Cl2(g) 2AlCl3(s) STEP 2Determine the moles of needed using a mole- RATIO 0.0556 mol Al x 3 mol Cl2= 0.0834 mol of Cl2 2 mol Al

  19. 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 Cl2)(0.0821 L• atm/mol K)(300. K) P 1.20 atm = 1.71 L of Cl2

  20. Learning Check What volume (L) of O2 at24 °C and 0.950 atm is needed to react with 28.0 g of NH3? 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

  21. Solution STEP 1Calculate the moles of given using molar mass or ideal gas law. 1 mol of NH3 = 17.03 g of NH3 1 mol NH3 and 17.03 g NH3 17.03 g NH3 Al 1 mol NH3 28.0 g NH3 x 1 mol NH3= 1.64 mol of NH3 17.03 g NH3

  22. Solution (continued) STEP 2Determine the moles of needed using a mole-mole factor. 5 mol of O2 = 4 mol of NH3 4 mol NH3and 5 mol O2 5 mol O2 4 mol NH3 1.64 mol NH3 x 5 mol O2 = 2.05 mol of O2 4 mol NH3

  23. Solution (continued) STEP 3Convert 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 O2 in the ideal gas law. V = nRT =(2.05 mol)(0.0821 L• atm/mol K)(297 K) P0.950 atm = 52.6 L of O2

  24. Learning Check What mass of Fe will react with 5.50 L of O2 atSTP? 4Fe(s) + 3O2(g) 2Fe2O3(s) 1) 13.7 g of Fe 2) 18.3 g of Fe 3) 419 g of Fe

  25. Solution STEP 1Calculate the moles of given using molar mass or ideal gas law. Use molar volume at STP to calculate moles of O2. 5.50 L O2 x 1 mol O2 = 0.246 mol of O2 22.4 L O2 STEP 2Determine the moles of needed using a mole-mole factor. 4 mol of Fe = 3 mol of O2 4 mol Feand 3 mol O2 3 mol O2 4 mol Fe 0.246 mol O2 x 4 mol Fe= 0.328 mol of Fe 3 mol O2

  26. Solution (continued) STEP 3Convert 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 O2 x 1 mol O2 x 4 mol Fex 55.85 g Fe = 18.3 g of Fe 22.4 L O2 3 mol O2 1 mol Fe

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