1. STOICHIOMETRY 4 Mole-Mole 4 Mole-Mass 4 Mass-Mass

2. 2 STOICHIOMETRYSTOICHIOMETRY - the study of the quantitative aspects of chemical reactions.

3. 3 STOICHIOMETRY CALCULATIONS Stoichiometric factor Moles reactant Moles product Coef. Want Coef. given

4. Mole -Mole Problems Moles of the Given Substance Moles of the Unknown Substance Conversion Factor: Coefficients from the balanced equation

5. Example Problem 2HCl + Fe = FeCl 2 + H 2 How many mole of HCl are need to react with 6 mole of Fe? 6 mol Fe X 2HCl 1 Fe = 12 mol HCl

6. Mole -Mass Problems (2 step) Moles of the Given Substance Mass of the given Substance Conversion Factor: Molar mass of the given X 

7. 7 STOICHIOMETRY CALCULATIONS Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Molar mass given Molar mass Unknown

8. Example Problem 2HCl + Fe = FeCl 2 + H 2 How many grams of HCl are need to react with 6 mole of Fe? 6 mol Fe X 2HCl 1 Fe = 12 mol HCl 12 mol HCl X 36 g HCl 1 mol = 432g

9. Mass -Mass Problems (3 step) Mass of the Given Substance Mass of the Unknown Substance Conversion Factors: Molar mass of each, and Coefficients from the balanced equation

10. 10 STOICHIOMETRY CALCULATIONS Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Molar mass given Molar mass Unknown

11. = 1 mol Fe 2HCl + Fe = FeCl 2 + H 2 How many grams of HCl are need to react with 56 g of Fe? 1 mol Fe X 2HCl 1 Fe = 2 mol HCl 2 mol HCl X 36 g HCl 1 mol = 72g HCl 1mol 56 g Fe 56 g of Fe X

12. 12 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

13. 13 454 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

14. 14 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. FACTOR.

15. 15 454 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

16. 16 454 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!

17. 17 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 5 How much N 2 O is formed? Total mass of reactants = total mass of products total mass of products 454 g NH 4 NO 3 = ___ g N 2 O + 204 g H 2 O mass of N 2 O = 250. g

18. 18 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield If you isolated only 131 g of N 2 O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields.

19. 19 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield

20. 20 PROBLEM: Using 5.00 g of H 2 O 2, what mass of O 2 and of H 2 O can be obtained? 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2 O 2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2