1. Lecture 109/21/05

2. Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Stoichiometric Calculations

3. Theoretical Yield vs. Percent Yield Maximum possible product % yield = actual / theoretical

4. P 4 (s) + 6 Cl 2 (g)  4PCl 3 (l) Initial1.00 mol6.00 mol0 mol Change (Δ) - 1.00 mol- 6.00 mol+4.00 mol Final 0 mol0 mol4.00 mol 1.00 mol of P 4 reacts with Cl 2 to form PCl 3. How many moles of Cl 2 are needed to react with the P 4 ? How many moles of PCl 3 are formed? Assumes complete reaction

5. What mass of Cl 2 is required to react completely with 1.45 g Phosphorous? What mass of PCl 3 is produced?

6. Step 1: Equation P 4 (s) + 6 Cl 2 (g)  4PCl 3 (l)

7. Step 2: Get moles Convert P 4 moles to mass

8. Step 3: Use stoichiometric factor Use factor to get moles of Cl 2

9. Step 4: Calculate mass from moles

10. 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

11. STEP 2 Convert mass reactant (454 g) to moles

12. STEP 3 Convert moles of reactant to moles of product using the stoichiometric factor = 11.4 mol H 2 O produced ALSO, 1:1 ratio of NH 4 NO 3 : N 2 O So 5.68 moles of each!

13. STEP 4 Convert moles of product to mass of product Total mass of reactants = total mass of products 454 g reactant = 454 g product

14. STEP 5 Calculate the percent yield If you actually only isolated 131 g of N 2 O, what is the percent yield?

15. The reactant that runs out 10 students ….8 desks Which is limiting? Limiting Reagent

16. If 365 g of CO is mixed with 65.0 g of H 2, what mass of CH 3 OH can be formed? Balanced equation CO (g) + 2 H 2 (g)  CH 3 OH (l)

17. Step 2: convert to moles CO: 12.7 moles H 2 :32.2 moles