1. Ch. 9-3 Limiting Reactants & Percent Yield

2. POINT > Define limiting reactant POINT > Identify which reactant is limiting in a reaction POINT > Define theoretical and actual yield POINT > Calculate percent yield based on results of a reaction

3. POINT > Define limiting reactant When one reactant is used up in a reaction, no more product can be formed. 2Cu + S  Cu 2 S The reactant that is depleted first is the limiting reactant. It “limits” the amount of product Other reactants that may still be present are excess reactants

4. 2Cu + S  Cu 2 S Ex. What is the limiting reactant when 80.0g Cu reacts with 25.0g S? The equation shows that 2 moles of Cu are needed for each 1 mole of S. The mole ratio of Cu / S is 2 /1 So, determine the number of moles from the masses given…

5. 2Cu + S  Cu 2 S 80.0g Cu = 1.26 mol Cu 25.0g S = 0.779 mol S If the Cu to S ratio is greater than 2/1 then Cu must be in excess and S must be the limiting reactant If the Cu to S ratio is less than 2/1 then S must be in excess and Cu must be the limiting reactant

6. 2Cu + S  Cu 2 S 80.0g Cu = 1.26 mol Cu 25.0g S = 0.779 mol S The ratio here is 1.62 / 1, so copper must be the limiting reactant When all the Cu has been used, excess S will remain

7. 4Al + 3O 2  2Al 2 O 3 What is the limiting reactant when 183.6g Al reacts with 153.6g O 2 ? 4 mol Al / 3 mol O 2 mole ratio is 1.33 6.80 mol Al and 4.80 mol O 2 6.80 4.80 = 1.41 There is excess Al, so the limiting reactant is O 2

8. Theoretical yield is the amount of product formed during a reaction as predicted using the chemical equation 4Al + 3O 2  2Al 2 O 3 Ex. If a reaction begins with 4 moles of Al and 3 moles of O 2 then the theoretical yield would be 2 moles of Al 2 O 3 which is 102g.

9. The actual yield is the amount of product actually produced during a lab reaction The actual yield is always less than the theoretical yield (for a variety of reasons)

10. Given eq. 2H 2 + O 2  2H 2 O If 8.0g of H 2 reacts with an excess of O 2 determine the theoretical yield of H 2 O in grams. 72g H 2 O

11. The percent yield is the ratio of actual yield to theoretical yield expressed as a percent Percent yield is a measure of reaction efficiency Percent Yield = Actual Yield Theoretical Yield x 100%

12. Ex. CaCO 3 (s)  CaO (s) + CO 2 (g) What is the theoretical yield of CaO if 24.8g CaCO 3 is heated? 24.8g CaCO 3  0.248mol CaCO 3 x = 0.248 mol CaO  Theoretical yield = 13.9g CaO ∆ 1 mol CaO 1 mol CaCO 3

13. CaCO 3 (s)  CaO (s) + CO 2 (g) Given a theoretical yield of 13.9g, what is the percent yield if 13.1g CaO is actually produced? 94.2% ∆

14. Read pages 296-302 Page 297 #1 Page 299 #1-2 Page 302 #1-2 FA page 302 #1-3