Percent Yield and Limited Reactants NC Essential Standard 2.2.4 Analyze the stoichiometric relationships inherent in a chemical reaction.

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Presentation transcript:

Percent Yield and Limited Reactants NC Essential Standard Analyze the stoichiometric relationships inherent in a chemical reaction

2 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) First write down the BALANCED equation! Now place the given numbers below the compounds.

3 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Two starting amounts? Where do we start? Simple mole to mole. We convert each to the ? Moles in separate calculations

4 Limiting/Excess/ Reactant and Theoretical Yield Problems Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O mol KO

5 Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O mol KO Based on: H 2 O = mol O mol H 2 O Limiting/Excess/ Reactant and Theoretical Yield Problems

6 Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O mol KO Based on: H 2 O = mol O mol H 2 O What is the theoretical yield? Always look for the smallest amount? This is always based upon the limiting reactant? Which in this case is KO 2 KO 2 has limited the amount of product. H 2 O = excess (XS) reactant!

7 Theoretical yield vs. Actual yield Suppose we use the theoretical yield form our last calculation mole O 2. In the actual experiment performed. Only 0.10 grams of product were recovered. Determine the % yield. Theoretical yield = g based on limiting reactant Actual yield = 0.1 g experimentally recovered

8 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) g47.0 g? g Based on: KO 2 = g O g KO Limiting/Excess Reactant Problem with % Yield

4 mol KO 2 1 mol O 2 3 mol O 2 1mol KO g KO 2 9 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) g 47.0 g ? g Based on: KO 2 = g O g KO Based on: H 2 O = g O 2 Question: If only 35.2 g of O 2 were recovered, what was the percent yield? 47.0 g H 2 O Limiting/Excess Reactant Problem with % Yield 32 g O 2 1 mol H 2 O 18 g H 2 O 1 mol O 2 3 mol O 2 2 mol H 2 O 32 g O 2