1. Limiting Reagent • 1. Limits or determines the amount of product that can be formed • 2. The reagent that is not used up is therefore the excess reagent • These types of problems require 2 sets of tracks. Quantities of both reagents will be given. Therefore, you need to find out which one is the limiting reagent.

2. Limiting Reagent • One track to determine limiting reagent • A second track to determine product

3. Limiting Reagent Example problem • How many grams of copper (I) Sulfide can be produced when 80.0 grams of Cu reacts with 25.0 grams of sulfur? • 2Cu + S --> Cu2S • Pick a reactant and calculate how much of the other reactant is needed. 80.0g Cu 1mol Cu 1mol S 32.1g S 63.5g Cu 2mol Cu 1mol S = 20.2g S So, 20.2 g of S is needed; 25.0g is supplied Plenty of S; therefore, Cu is limiting reagent. Use Cu to solve the problem 80.0g Cu 1mol Cu 1mol Cu2S 159.1g Cu2S 63.5g Cu 2mol Cu 1mol Cu2S = 1.00x102 g Cu2S

4. Limiting Reagent Example Problem - Your Turn • How many grams of hydrogen can be produced when 5.00g of Mg is added to 6.00 g of HCl? • Mg + 2 HCl --> MgCl2 + H2 • Pick a reactant and calculate how much of the other reactant is needed. 5.00g Mg 1mol Mg 2mol HCl 36.5g HCl 24.3g Mg 1 mol Mg 1mol HCl = 15.0g HCl Need 15.0g HCl; have 6.00 g HCl Not enough HCl; therefore, HCl is limiting reagent Use HCl to solve the problem 6.00g HCl 1mol HCl 1mol H2 2.0g H2 36.5g HCl 2mol HCl 1mol H2 = 0.164 g H2

5. Limiting Reagent Example problem- Your Turn • Acetylene (C2H2) will burn in the presence of oxygen. How many grams of water can be produced by the reaction of 2.40 mol of acetylene with 7.4 mol of oxygen? • 2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O • Pick a reactant and calculate how much of the other reactant is needed • 2.40 mol C2H2 5 mol O2 • 2 mol C2H2 = 6.00 mol O2 • Need 6.00mol O2; have 7.4mol O2 • Plenty of O2; so, C2H2 is L.R. • 2.4mol C2H2 2mol H2O 18.0g H2O • 2mol C2H2 1mol H2O2 • = 43.2 g H2O