Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 9 Chemical Quantities in Reactions 9.3 Limiting Reactants A ceramic brake disc in a sports car withstands temperature of 1400°C.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Reacting Amounts In a table setting, there are 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 6 forks, 4 spoons, and 7 knives? What is the limiting item?

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Reacting Amounts Four table settings are possible. Initially Used Extra forks 6 42 spoons knives The limiting item is the spoon.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 Example of Everyday Limiting Reactant How many peanut butter sandwiches can be made from 8 slices of bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches can be made. The bread is the limiting reactant.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 Example of Everyday Limiting Reactant How many peanut butter sandwiches can be made from 8 slices of bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich can be made. The peanut butter is the limiting item.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Limiting Reactants When 4.00 mol of H 2 is mixed with 2.00 mol of Cl 2, how many moles of HCl can form? H 2 (g) + Cl 2 (g)  2HCl (g) 4.00 mol 2.00 mol ??? Mol

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 Limiting Reactants Using Moles Calculate the moles of product from each reactant, H 2 and Cl 2. moles of HCl from moles of H mol H 2 x 2 mol HCl = 8.00 mol of HCl 1 mol H 2 (not possible) moles of HCl from moles of Cl mol Cl 2 x 2 mol HCl = 4.00 mol of HCl 1 mol Cl 2 (smaller number) The limiting reactant is Cl 2 because it produces the smaller number of moles of HCl.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 Checking Calculations Initial Reactants H mol Cl mol Product 2HCl 0 mol React/ Form  2.00 mol mol Left after reaction 2.00 mol Excess 0 mol Limiting 4.00 mol

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 Calculating Mass of Product from a Limiting Reactant

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g of H 2 and 24.0 g of O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 Limiting Reactants Using Mass STEP 1 Use molar mass to convert the grams of each reactant to moles. Given 8.00 g of H 2 and 24.0 g of O 2 Need grams of H 2 O 8.00 g H 2 x 1 mol H 2 = 3.97 mol of H g H g O 2 x 1 mol O 2 = mol of O g O 2

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 Limiting Reactants Using Mass (continued) STEP 2 Write mole-mole factors using the coefficients in the equation. 1 mol of O 2 = 2 mol of H 2 O 1 mol O 2 and 2 mol H 2 O 2 mol H 2 O 1 mol O 2 2 mol of H 2 = 2 mol of H 2 O 2 mol H 2 and 2 mol H 2 O 2 mol H 2 O 2 mol H 2

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 Limiting Reactants Using Mass (continued) STEP 3 Calculate moles of product from each reactant and determine the limiting reactant. moles of H 2 O from moles of H mol H 2 x 2 mol H 2 O = 3.97 mol of H 2 O 2 mol H 2 (not possible) moles of H 2 O from moles of O mol O 2 x 2 mol H 2 O = 1.50 mol of H 2 O 1 mol O 2 (smaller number) The limiting reactant is O 2 because it produces the smaller moles (1.50 mol) of H 2 O.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 Limiting Reactants Using Mass (continued) STEP 4 Determine the moles of product or calculate grams of product using molar mass mol H 2 O x g H 2 O = 27.0 g of H 2 O 1 mol H 2 O

Basic Chemistry Copyright © 2011 Pearson Education, Inc. Limiting 16