1. Limiting Reactant II and Percent Yield A.K.A. Stoichiometry

2.  The reactant that limits the amount of product produced  Limiting reactant is consumed fully in a chemical reaction  Excess reactant remains in a chemical reaction What is a Limiting Reactant?

3. Remember The Sundae Example… The chocolate syrup was consumed fully. The ice cream and cherries are left over.

4.  Calculate using mass-mass conversion; find which reactant produces the least amount of product  Or use mole-mole conversion to determine which reactant is consumed first How do we determine which reactant is the limiting reactant?

5.  What is the limiting reactant when 10.0 g of SiO 2 react with 5.0 g of HF?  Create a conversion pathway using mass-mass conversion SiO 2 (s) + 4 HF(l) → SiF 4 (g) + 2 H 2 O(l)

6. Mass of SiO 2 Molar Mass of SiO 2 Mole RatioMolar Mass of H 2 O Mass of H 2 O 10.0 g SiO 2 1 mol SiO 2 2 mol H 2 O18.02 g H 2 O= 0.166 g H 2 O 60.086 g SiO 2 1 mol SiO 2 1 mol H 2 O SiO 2 (s) + 4 HF(l) → SiF 4 (g) + 2 H 2 O(l) Mass of HFMolar Mass of HF Mole RatioMolar Mass of H 2 O Mass of H 2 O 5.0 g HF1 mol HF2 mol H 2 O18.02 g H 2 O= 0.263 g H 2 O 20.008 g HF4 mol HF1 mol H 2 O SiO 2 is the limiting reactant!

7. Limiting Reactant Relay Time for

8.  How efficient is a chemical reaction?  Does the reaction go to completion?  How much product is produced? How can we predict the amount of product produced?

9. How Does Sundae Production and Percent Yield Relate? If you made only 25 sundaes but You really needed 40, what was your production yield? Actual yield = 25 sundaes Production (theoretical) yield = 40 sundaes Percent yield = 25 x 100 % = 63% 40

10.  5.00 g of Cu is mixed with an excess of AgNO 3.  The reaction produces 15.2 g of Ag  What is the percent yield for this reaction? We want to know how much product is produced?

11.  Create your conversion pathway using mass-mass conversion Cu + 2 AgNO 3  2 Ag + Cu(NO 3 ) 2 Mass of CuMolar Mass of Cu Mole RatioMolar Mass of Ag Mass of Ag 5.00 g Cu1 mol Cu2 mol Ag107.9 g Ag= 17.0 g Ag 63.5 g Cu1 mol Cu1 mol Ag 17.0 g Ag is our theoretical yield; need to use it to calculate percent yield

12. Percent Yield = Actual yieldx 100 Theoretical yield Actual/Theoretic al PercentPercent Yield = 15.2 g Agx 100= 89.4 % 17.0 g Ag

13. Percent Yield Worksheet Your turn…

14.  What is the percent yield when 24.8 g of CaCO 3 decomposes to give 13.1 g CaO?  CaCO 3  CaO + CO 2  Plan your conversion pathway  Utilize mass-mass conversion One more example

15. Mass of CaCO 3 Molar Mass of CaCO 3 Mole RatioMolar Mass of CaO Mass of CaO 24.8 g CaCO 3 1 mol CaCO 3 1 mol CaO56.1 g CaO= 13.9 g CaO 100.1 g CaCO 3 1 mol CaCO 3 1 mol CaO CaCO 3  CaO + CO 2 The theoretical yield is 13.9 g of CaO; what is our percent yield? The reaction made 13.1 g CaO Actual/Theoreti cal PercentPercent Yield = 13.1 gx 100= 94.2 % 13.9 g

16. Exit Ticket Time to fill out an

17.  Read over Stoichiometry Lab carefully  Answer pre-laboratory questions Homework