1. Limiting Reactants, Theoretical Yield, and % Yield

2. Limiting Reactants Limiting reactants are reactants that limit the amount of products that can be formed –They are consumed first in the reaction Excess reactants are reactants that are not completely consumed –They are left over

3. Limiting Reactant Problems A limiting reactant problem can be easily identified because you have TWO givens You basically are just going to do 2 stoichiometry problems

4. Steps to determine the limiting reactant 1. Balance the equation 2. Calculate the number of moles of each reactant 3. Convert one given to moles of 1 product (you can pick any product) 4. Repeat for second given 5. The reactant that produces the smallest amount (moles) of product is the LR 6. Use the LR in any calculations

5. Example If 18.1 g or NH 3 are reacted with 90.4 g of CuO, which is the limiting reactant and how many grams of N 2 will be formed? NH 3 + CuO  N 2 + Cu + H 2 O Balance the equation 2NH 3 + 3CuO  N 2 + 3Cu + 3H 2 O

6. Determine the number of moles 18.1 g NH 3 x 1 mol NH 3 x 1 mol N 2 = 0.53 mol N 2 17.03 g NH 3 2 mol NH 3 17.03 g NH 3 2 mol NH 3 90.4 g CuO x 1 mol CuO x 1 mol N 2 = 0.38 mol N 2 79.55 g CuO 3 mol CuO 79.55 g CuO 3 mol CuO CuO produces less product, so it is the LR Example

7. CuO is what we use to calculate the amount of N 2 produced 0.38 mol N 2 x 28.0 g N 2 = 10.6 g N 2 1 mol N 2 1 mol N 2 Example N 2 10.6 g N 2

8. Another Example If 25.0 kg of nitrogen are reacted with 5.00 kg of hydrogen to form ammonia, how much ammonia will be produced? Balanced equation N 2 + 3H 2  2NH 3 28.0 kg NH 3

9. Theoretical Yield The theoretical yield is the amount of product that would be produced if all of the limiting reactant were completely consumed –The maximum amount of product that can be produced with the given quantities of reactants

10. Percent Yield The percent yield is the actual yield of product, expressed as a percent Actual yield x 100% = percent yield Actual yield x 100% = percent yield Theoretical yield Theoretical yield When performing an experiment, things do not always go exactly perfect Some product may get spilled, some may get sneezed on, or the reaction may not have gone to completion

11. Example When 68.5 kg CO react with 8.60 kg H 2, what is the percent yield if 35.7 kg of methanol (CH3OH)are formed? Balanced equation 2H 2 + CO  CH 3 OH

12. Example Find the LR 68.5 kg CO x 1000 g CO x 1 mol CO x 1 mol CH 3 OH = 2440 mol CH 3 OH 1 kg CO 28 g CO 1 mol CO 1 kg CO 28 g CO 1 mol CO 8.60 kg H 2 x 1000 g H 2 x 1 mol H 2 x 1 mol CH 3 OH = 2135 mol CH 3 OH 1 kg H 2 2 g H 2 2 mol H 2 1 kg H 2 2 g H 2 2 mol H 2 H 2 produces less CH 3 OH, so it is the LR H 2 produces less CH 3 OH, so it is the LR

13. Example If all of the H 2 is consumed, 2135 mol CH 3 OH will be produced 2135 mol CH 3 OH x 32 g CH 3 OH x 1 kg CH 3 OH = 68.6 kg CH 3 OH 1 mol CH 3 OH 1000 g CH 3 OH 1 mol CH 3 OH 1000 g CH 3 OH theoretical yield theoretical yield

14. Example Calculate the %yield 35.7 kg CH 3 OH x 100% = 52 % 68.6 kg CH 3 OH 68.6 kg CH 3 OH

15. Another example C 7 H 6 O 3 + C 4 H 6 O 3  C 9 H 8 O 4 + HC 2 H 3 O 2 When 1.50 g C 7 H 6 O 3 with 2.00 g C 4 H 6 O 3, 1.50 g C 9 H 8 O 4 was produced. What is the %yield for this experiment? 76.5%