1. Limiting Reactants

2. Limiting vs. Excess  Limiting Reactant-  Excess Reactant- The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed. The reactant in a chemical reaction that remains when the reaction stops. The excess reactant remains because there is nothing left with which it can react.

3. Real Life Example…  If there are only 8 car bodies, then only 8 cars can be made. Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up. Limiting Reactant Excess Reactant

4. Finding the Limiting Reactant  Find the number of moles of one product for each reactant. The reactant that produces the smallest amount of product is the limiting reactant. involves doing a one or two step stoichiometry problem.

5. Example Problem #1  A 2.00 mol sample of ammonia is mixed with 4.00 mol of oxygen. Which is the limiting reactant and which is in excess? 4 NH 3 + 5 O 2  4 NO + 6 H 2 O

6. 5 mol O 2 4.00 mol O 2 4 mol NH 3 2.00 mol NH 3 4 mol NO  4 NH 3 + 5 O 2  4 NO + 6 H 2 O = 2.00 mol NO = 3.20 mol NO NH 3 is Limiting Reactant O 2 is Excess Reactant

7. Finding the Amount of Product  Phase 1: Calculate the number of moles of one product for all reactants.  Phase 2: Use limiting reactant to calculate the amount of product(s) involves doing a one or two step stoichiometry problem for each product you need to find

8. Example Problem #2  A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. What is the mass of nitrogen monoxide & water produced?

9. 5 mol O 2 1 mol O 2 32.0 g O 2 4.00 g O 2 4 mol NH 3 1 mol NH 3 17.0 g NH 3 2.00 g NH 3 4 mol NO  4 NH 3 + 5 O 2  4 NO(g) + 6 H 2 O Phase 1: Solve for moles of one product for ALL reactants = 0.117 mol NO = 0.100 mol NO O 2 is Limiting Reactant NH 3 is Excess Reactant

10. 1 mol H 2 O4 mol NO 0.100 mol NO 1 mol NO 0.100 mol NO30.0 g NO If needed… Phase 2: Solve for mass of products = 3.00 g NO produced 6 mol H 2 O18.0 g H 2 O = 2.70 g H 2 O produced Once you determine the limiting reactant, use that to solve any other problems. Finding mass of NO: From step 1 O 2 was L.R., so start with the answer from O 2 Finding mass of H 2 O: From step 1 O 2 was L.R., so start with the answer from O 2

11. Example Problem #3  When 0.65 moles of oxygen reacts with 0.56 moles of potassium how many grams of potassium oxide will be produced?  4 K + O 2  2 K 2 O 1 mol O 2 0.65 mol O 2 2 mol K 2 O = 1.3 mol K 2 O 4 mol K 0.56 mol K2 mol K 2 O = 0.28 mol K 2 O K is Limiting Reactant O 2 is Excess Reactant 1 mol K 2 O 0.28 mol K 2 O94.2 g K 2 O = 26 g K 2 O produced Phase 1: Phase 2:

12. Additional Examples 4. How much CO 2 forms when 3.7 moles of C 6 H 6 reacts with 28.4 moles of oxygen. 5. How much Al 2 O 3 will form if 21.0 g of Al is reacted with 50.0 g of Fe 2 O 3 ? 2C 6 H 6 + 15O 2 → 12CO 2 + 6H 2 O 2Al + Fe 2 O 3 → Al 2 O 3 + 2Fe

13. Additional Examples 6. If 15.3 g of silver nitrate reacts with 28.4 g of zinc chloride, what mass of silver chloride is produced? 2AgNO 3 + ZnCl 2 → 2AgCl + Zn(NO 3 ) 2