UNIT VII Excess & Percentage Yield Unit 7: Lesson 2

L IMITING REAGENTS & E XCESS AMOUNTS SO far, we’ve only considered reactions assuming that ALL of the reactants will be USED UP entirely so zero is left at the end of the reaction. Wellllll, this is not always true! WHAT to do?!!?

VII.4 STOICHIOMETRY OF EXCESS QUANTITIES Often in a reaction, not all of the reactants are used up – frequently one or more reactants are in excess LIMITING REAGENT : is the reactant you run out of first! So it limits how much product you can make. EXCESS REAGENT: is the one you have left over!

VII.4 STOICHIOMETRY OF EXCESS QUANTITIES Ex: you are given 12 toasts of bread, a gallon of mustard and 3 pieces of ham – how many sandwiches can you make??? Limiting reagent = Excess reagents = The limiting reagent determines how much product you can make

STOICHIOMETRY OF EXCESS QUANTITIES Well, how do we find out the limiting reagent in a chemical reaction!? In order to FIND OUT which reactant is limiting: 1- start with the amount of reactants you are given 2- change each reactant  moles. 3- find out how many moles of products you would get. 4- the reactant that makes the lowest number of moles is your limiting reagent! YOUR LIMITING REAGENT IS THE ONE YOU USE TO PREDICT HOW MUCH PREDICT YOU WILL HAVE!

VII.4 STOICHIOMETRY OF EXCESS QUANTITIES Ex: #1 Copper reacts with sulfur to form copper (I) sulfide. If 10.6g of copper reacts with 3.83g of S how much product will be formed?

Ex: #2 Mg (s) + 2HCl (g) → MgCl 2(s) + H 2(g) If 10.1 mol of magnesium and 4.87 mol of HCl gas are reacted, how many moles of gas will be produced? How much excess reagent remains?

TRY: If 10.3 g of aluminum are reacted with 51.7g of CuSO 4 how much copper will be produced? How much excess reagent will remain?

Ex: #3 When 79.1 g of zinc are reacted with 1.05L of 2.00M HCl, to produce ZnCl 2, and H 2 gas, which reactant will be in excess and by how much?

PERCENTAGE YIELD Percentage Yield: is used to describe the amount of product actually obtained as a percentage of the expected amount. 2 reasons: 1. reactants may not all react 2. some products are lost Yield : is the amount made in a chemical reaction

Actual Yield – what you ACTUALLY get in the lab when the chemicals are mixed Theoretical Yield – what the balanced equation says you should make Percent Yield - Actual Yield (obtained) x 100% Theoretical Yield (expected) Percent purity- pure x 100% impure THREE TYPES of YIELD

Ex: #1 When 45.8g of potassium carbonate, K 2 CO 3 are reacted completely with an excess of HCl, 46.3 g of KCl are produced. Water and CO 2 also are formed. Calculate the theoretical yield and the percent yield of KCl.

Ex: #2 2Al + 3CuSO 4 → Al 2 (SO 4 ) 3 + 3Cu When 3.92 g of Al are reacted with excess copper(II) sulfate, 6.78 g of copper is produced. a) What is the actual yield?

Ex: #2 (continued) 2Al + 3CuSO 4 → Al 2 (SO 4 ) 3 + 3Cu When 3.92 g of Al are reacted with excess copper(II) sulfate, 6.78 g of copper is produced. b) What is the theoretical yield?

Ex: #2 (continued) 2Al + 3CuSO 4 → Al 2 (SO 4 ) 3 + 3Cu When 3.92 g of Al are reacted with excess copper(II) sulfate, 6.78 g of copper is produced. c) What is the percentage yield of Cu?

Ex: #2 (continued) 2Al + 3CuSO 4 → Al 2 (SO 4 ) 3 + 3Cu d) If you started with 9.73g of Al as your L.R, how much copper would you expect?

End of practicing theoretical yields and percentage yields

H OW TO CALCULATE ACTUAL YIELD What if you are not given the actual yield? Instead, you have a percentage (%) How to solve it? There are two possible situations: Actual yield of a product Actual yield of a reactant

Actual yield of a product Actual amount of PRODUCTS FOUND is LOWER than the value expected. To solve: calculate T.Y using stoichiometry then multiply (x) “T.Y” by the decimal equivalent of the % yield or % purity For actual yield of products: A.Y= T.Y x %yield

Actual yield of a reactant Actual amount of REACTANTS NEEDED is GREATER than the value expected To solve: calculate the “T.Y” using stoichiometry then divide (÷) “T.Y” by the decimal equivalent of % yield or % purity For actual yield of reactants: A.Y= T.Y / %yield

Ex: What actual mass of Fe 2 O 3 is produced when 2.30 g of FeCO 3 are reacted with an excess of O 2 according to the reaction: 4FeCO 3 + O 2 → 2Fe 2 O 3 + 4CO 2, if the reaction has a 67.0 % yield?

Ex: What actual mass of CO 2 is required to make 8.00g of K 2 CO 3(s) according to the reaction: 4KO 2(s) + 2CO 2(g) → 2K 2 CO 3(s) + 3O 2(g) if the reaction has a 85.0% yield?

HOMEWORK Excess amounts: p.133 # % questions: p. 137 # UNIT 7 STOICHIOMETRY : FINITOOOOOOO!