Presentation on theme: "Limiting Reactants. Limiting vs. Excess Limiting Reactant- Excess Reactant- The reactant in a chemical reaction that limits the amount of product."— Presentation transcript:
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.
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
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.
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 O 2 4 NO + 6 H 2 O
5 mol O mol O 2 4 mol NH mol NH 3 4 mol NO 4 NH 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
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
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?
5 mol O 2 1 mol O g O g O 2 4 mol NH 3 1 mol NH g NH g NH 3 4 mol NO 4 NH O 2 4 NO(g) + 6 H 2 O Phase 1: Solve for moles of one product for ALL reactants = mol NO = mol NO O 2 is Limiting Reactant NH 3 is Excess Reactant
1 mol H 2 O4 mol NO mol NO 1 mol NO 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
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 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:
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 O 2 → 12CO 2 + 6H 2 O 2Al + Fe 2 O 3 → Al 2 O 3 + 2Fe
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