1 Chapter 12: Day 5 Ch12_stoic

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

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

4 454 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 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.

6 454 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

7 454 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!

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

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 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 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

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12

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

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

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)

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 = g of Al 1 mol Al and g Al g Al 1 mol Al 1.50 g Al x 1 mol Al = mol of Al g Al

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 mol Al x 3 mol Cl 2 = mol of Cl 2 2 mol Al

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 = 300. K V = nRT = ( mol Cl 2 )( L atm/mol K)(300. K) P 1.20 atm = 1.71 L of Cl 2

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 21 What volume (L) of O 2 at 24 °C and 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

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 = g of NH 3 1 mol NH 3 and g NH g NH 3 Al 1 mol NH g NH 3 x 1 mol NH 3 = 1.64 mol of NH g NH 3 Solution

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 mol NH 3 x 5 mol O 2 = 2.05 mol of O 2 4 mol NH 3 Solution (continued)

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 = 297 K Place the moles of O 2 in the ideal gas law. V = nRT =(2.05 mol)( L atm/mol K)(297 K) P atm = 52.6 L of O 2 Solution (continued)

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

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 L O 2 x 1 mol O 2 = mol of O 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 mol O 2 x 4 mol Fe = mol of Fe 3 mol O 2 Solution

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 = g of Fe 1 mol Fe and g Fe g Fe 1 mol Fe mol Fe x 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 g Fe = 18.3 g of Fe 22.4 L O 2 3 mol O 2 1 mol Fe Solution (continued)