1. Bell Ringer When 6.58 g SO 3 and 1.64 g H 2 O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield? SO 3 +H2OH2OH 2 SO 4 6.58 g SO 3 x 80.07 g SO 3 1 mol SO 3 x 1 mol H 2 SO 4 x 98.09 g H 2 SO 4 = 8.06 g H 2 SO 4 1.64 g H 2 O x 18.02 g H 2 O 1 mol H 2 O x 1 mol H 2 SO 4 x 98.09 g H 2 SO 4 = 8.93 g H 2 SO 4

2. Bell Ringer When 6.58 g SO 3 and 1.64 g H 2 O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield? SO 3 +H2OH2OH 2 SO 4 Info we’ve learned:Theoretical Yield = 8.06 g H 2 SO 4 % Yield = Actual Yield Theoretical Yield x100 % = 7.99 g H 2 SO 4 8.06 g H 2 SO 4 99.1 %

3. Limiting Factors & Percent Yield Quiz

4. Homework Answers 1.D 2.A 3.C 4.B 5.A 6.C 7.C 8.A 9.B 10.D 11. Mass-mass 12. Mass-volume 13. Mole-mole 14. Limiting reactant 15. Volume-volume 17. 0.52 mol PBr 3 19. 0.13 mol Na 21. 50 g NaClO 3 23. 5000 g HCl 25. 4770 g H 2 O 27. 98 g AgCl; 120 g AgNO 3 29. 700 g CO 2 ; 500 g O 2 31. 89.6 L H 2 33. 234 g ZnSO 4 35. 7.75 L O 2 39. 8.06 g H 2 SO 4 ; 99.1 %

5. Stoichiometry Review Ms. Besal 3/7/2006

6. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

7. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

8. Mole-Mole Problems 1 conversion step –Given: moles “A” –Required: moles “B” Convert moles “A” to moles “B” using mole ratio. The mole ratio is used in EVERY STOICHIOMETRY PROBLEM. EVER. I PROMISE. 2 H 2 + O 2 2 H 2 O How many moles of water can be formed from 0.5 mol H 2 ? 0.5 mol H 2 x 2 mol H 2 2 mol H 2 O = 0.5 mol H 2 O

9. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

10. Mass-Mole Problems 2 conversion steps –Given: mass “A” –Required: moles “B” Step 1: convert grams “A” to moles “A” using Periodic Table Step 2: convert moles “A” to moles “B” using mole ratio 2 H 2 + O 2 2 H 2 O How many moles of water can be formed from 48.0 g O 2 ? 48.0 g O 2 x 1 mol O 2 2 mol H 2 O = 3.00 mol H 2 O 32.00 g O 2 1 mol O 2 x

11. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

12. Mass-Mass Problems 3 conversion steps –Given: mass “A” –Required: mass “B” Step 1: convert grams “A” to moles “A” using Periodic Table Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to grams “B” using Periodic Table 2 H 2 + O 2 2 H 2 O How many grams of water can be formed from 48.0 g O 2 ? 48.0 g O 2 x 1 mol O 2 2 mol H 2 O = 32.00 g O 2 1 mol O 2 xx 18.02 g H 2 O 1 mol H 2 O 54.1 g H 2 O

13. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

14. Mass-Volume Problems 3 – 4 conversion steps –Given: mass “A” –Required: volume “B” Step 1: convert grams “A” to moles “A” using Periodic Table Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to liters “B” 2 H 2 + O 2 2 H 2 O 48.0 g H 2 O x 2 mol H 2 O 1 mol O 2 = 18.02 g H 2 O 1 mol H 2 O xx 22.4 L O 2 1 mol O 2 How many liters of oxygen are necessary to create 48.0 g H 2 O? 29.8 L O 2

15. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

16. Volume-Mass Problems 3 – 4 conversion steps –Given: volume “A” –Required: mass “B” Step 1: convert liters “A” to moles “A” Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to grams “B” using Periodic Table 2 H 2 + O 2 2 H 2 O 36.0 L O 2 x 1 mol O 2 2 mol H 2 O = 22.4L O 2 1 mol O 2 xx 18.02 g H 2 O 1 mol H 2 O How many grams of water are formed by reacting 36.0 L O 2 ? 58.7 g H 2 O

17. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

18. Volume-Volume Problems 3 – 5 conversion steps –Given: volume “A” –Required: volume “B” Step 1: convert liters “A” to moles “A” Step 2: convert moles “A” to moles “B” using mole ratio Step 3: convert moles “B” to liters “B” 2 H 2 + O 2 2 H 2 O 5.0 L O 2 x 1 mol O 2 2 mol H 2 = 22.4 L O 2 1 mol O 2 xx 22.4 L H 2 1 mol H 2 How many liters of H 2 are required to react with 5.0 L O 2 ? 10. L H 2

19. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

20. Limiting Reactant Problems Quantities are given for each reactant. 2 parallel equations Solve each equation for product desired and determine limiting reactant. Use Limiting Reactant to solve for amount or excess reactant used. Subtract amount excess reactant used from amount given to determine how much is left over.

21. Limiting Reactant Problems If you start with 10.0 g of O 2 and 5.00 g H 2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be? 2 H 2 +O2O2 2 H 2 O 10.0 g O 2 5.00 g H 2 x x 2.02 g H 2 1 mol H 2 1 mol O 2 32.00 g O 2 2 mol H 2 O 1 mol O 2 2 mol H 2 x x= =x 1 mol H 2 O 18.02 g H 2 O 11.3 g H 2 O x 1 mol H 2 O 18.02 g H 2 O 44.06 g H 2 O THEORETICAL YIELD LIMITING REACTANT EXCESS REACTANT

22. Limiting Reactant Problems Info we know so far: Limiting Reactant = O 2 Excess Reactant = H 2 10.0 g O 2 x x 1.26 g H 2 USED 5.00 g H 2 – 1.26 g H 2 =3.74 g H 2 LEFT OVER 1 mol O 2 2 mol H 2 x 2 H 2 +O2O2 2 H 2 O If you start with 10.0 g of O 2 and 5.00 g H 2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be? 1 mol O 2 32.00 g O 2 2.02 g H 2 1 mol H 2 =

23. Types of Stoichiometry Problems Mole-Mole Mass-Mole Mass-Mass Mass-Volume Volume-Mass Volume-Volume Limiting Reactant Percent Yield

24. Percent Yield Problems Critical Information: –Theoretical Yield –Actual Yield –Percent Yield You will be given one of these May or may not be given 2 H 2 +O2O2 2 H 2 O Determine the actual yield of a reaction between 6.25 g H 2 and excess O 2 that has a 85% percent yield. 6.25 g H 2 x 2.02 g H 2 1 mol H 2 2 mol H 2 O 2 mol H 2 x=x 1 mol H 2 O 18.02 g H 2 O 55.6 g H 2 O 85 % = x 100 % 55.6 g H 2 O ? ACTUAL YIELD = 47.3 g H 2 O THEORETICAL YIELD

25. Homework Answers 1.D 2.A 3.C 4.B 5.A 6.C 7.C 8.A 9.B 10.D 11. Mass-mass 12. Mass-volume 13. Mole-mole 14. Limiting reactant 15. Volume-volume 17. 0.52 mol PBr 3 19. 0.13 mol Na 21. 50 g NaClO 3 23. 5000 g HCl 25. 4770 g H 2 O 27. 98 g AgCl; 120 g AgNO 3 29. 700 g CO 2 ; 500 g O 2 31. 89.6 L H 2 33. 234 g ZnSO 4 35. 7.75 L O 2 39. 8.06 g H 2 SO 4 ; 99.1 %