1. Limiting Reactant Stoichiometry

2. 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

3. Example: If you had 152. 5 g of CO and 24
Example: If you had g of CO and 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

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

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

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

7. 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 = mol CH3OH
mol CO
5.446 mol CH3OH x 32.0 g = g CH3OH
mol