1 Chapter 8 Chemical Quantities in Reactions 8.3 Limiting Reactants Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 Limiting Reactant A limiting reactant in a chemical reaction is the substance that Is used up first. Stops the reaction. Limits the amount of product that can form.

3 Reacting Amounts In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

4 Reacting Amounts Four table settings can be made. Initially Use Left over plates 5 41 forks 6 42 spoons knives The limiting item is the spoon.

5 Example of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item. Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

6 Example of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item. Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

7 Limiting Reactants When 4.00 mol H 2 is mixed with 2.00 mol Cl 2,how many moles of HCl can form? H 2 (g) + Cl(g)  2HCl (g) 4.00 mol 2.00 mol ??? mol Calculate the moles of product from each reactant, H 2 and Cl 2. The limiting reactant is the one that produces the smaller amount of product.

8 Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

9 Limiting Reactants Using Moles HCl from H mol H 2 x 2 mol HCl = 8.00 mol HCl 1 mol H 2 (not possible) HCl from Cl mol Cl 2 x 2 mol HCl = 4.00 mol HCl 1 mol Cl 2 (smaller number) The limiting reactant is Cl 2 because it is used up first. Thus Cl 2 produces the smaller number of moles of HCl.

10 Checking Calculations InitiallyH mol Cl mol 2HCl 0 mol Reacted/ Formed mol mol Left after reaction 2.00 mol Excess 0 mol Limiting 4.00 mol

11 Limiting Reactants Using Mass If 4.80 mol Ca mixed with 2.00 mol N 2, which is the limiting reactant? 3Ca(s) + N 2 (g)  Ca 3 N 2 (s) Moles of Ca 3 H 2 from Ca 4.80 mol Ca x 1 mol Ca 3 N 2 = 1.60 mol Ca 3 N 2 3 mol Ca (Ca used up) Moles of Ca 3 H 2 from N mol N 2 x 1 mol Ca 3 N 2 = 2.00 mol Ca 3 N 2 1 mol N 2 (not possible) All Ca is used up when 1.60 mol Ca 3 N 2 forms. Thus, Ca is the limiting reactant.

12 Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l)

13 Limiting Reactants Using Mass Calculate the grams of H 2 for each reactant. H 2 : 8.00 g H 2 x 1 mol H 2 x 2 mol H 2 O x g H 2 O g H 2 2 mol H 2 1 mol H 2 O = 71.5 g H 2 O (not possible) O 2 : 24.0 g O 2 x 1 mol O 2 x 2 mol H 2 O x g H 2 O g O 2 1 mol O 2 1 mol H 2 O = 27.0 g H 2 O (smaller) O 2 is the limiting reactant.