Limiting Reactant Stoichiometry

limiting reactant: reactant that is consumed completely in a reaction; determines the amount of product made excess reactant: reactant that will not be used up in a reaction

Example: If you had 152. 5 g of CO and 24 Example: If you had 152.5 g of CO and 24.50 g of H2, what mass of CH3OH could be produced? CO + 2H2  CH3OH Step 1: Determine if it is a limiting reactant problem. If the amounts of two or more reactants are given, it is a limiting reactant problem

Step 2: How many moles of each reactant do you have? 152.5 g CO x 1 mole = 5.446 mol CO 28.0 g 24.50 g H2 x 1 mole = 12.25 mol H2 2.0 g

Step 3: Divide the number of moles of each by its coefficient in the equation. 5.446 mol CO = 5.446 12.25 mol H2 = 6.125 1 2 *This is only a quick way to determine the limiting reactant.

Step 4: After step 3, the reactant with the smaller number is the limiting reactant. CO is the limiting reactant because 5.446 is smaller than 6.125.

Step 5: Use the number of moles of the limiting reactant and stoichiometry to predict how much product can be produced. 5.446 mol CO x 1 mol CH3OH = 5.446 mol CH3OH 1 1 mol CO 5.446 mol CH3OH x 32.0 g = 174.3 g CH3OH 1 1 mol