1. Limiting Reactant

2.  Determine which reactant is left over in a reaction.  Identify the limiting reactant and calculate the mass of the product.  Calculate the amount of excess reactants.

3. How many BLT Sandwiches can you make with 3 Tomatoes, 2 Kg of Bacon, 1 head of lettuce and 4 slices of bread? 2

4. Limiting Reactant - determines the amount of product that can be formed in a reaction. 2 moles + 3 moles Reactants remaining are called the excess reactants. H H H N H H N N H H H H H H H N N 2 (g) + 3 H 2 (g) 2 NH 3 (g) N N

5. Limiting Reactant Problems Step 1: Determine which reactant will be used up first – pick one and calculate the other. (limiting reactant) Step 2: Use the limiting reactant to determine the amount of product. Step 3: Calculate the amount of reactant left over from the moles of limiting reactant used.

6. How much NaCl is produced and what is left over when 6.70 mol of Na react with 3.20 mol of Cl 2 ? 2 Na (s) + Cl 2 (g) 2 NaCl (s) Cl 2 - limiting reactant, Na - excess reactant 1 mol Cl 2 2 mol Na 6.70 mol Na = 3.35 mol Cl 2 = 6.40 mol NaCl 1 mol Cl 2 2 mol NaCl 3.20 mol Cl 2

7. = 6.40 mol Na 1 mol Cl 2 2 mol Na 3.20 mol Cl 2 6.70 mol - 6.40 mol = 0.30 mol Na excess 2 Na (s) + Cl 2 (g) 2 NaCl (s) …when 6.70 mol of Na react with 3.20 mol of Cl 2 ? Cl 2 - limiting reactant, Na - excess reactant

8. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H 2 - limiting reactant, N 2 - excess reactant How many grams of ammonia can be made from 3.50 g of H 2 gas and 18.0 g of nitrogen gas? What’s left? 1 mole N 2 3 mole H 2 1 mole H 2 2.0 g H 2 3.50 g H 2 28.0 g N 2 1 mole N 2 = 16.3 g N 2

9. 18.0 g – 16.3 g = 1.68g N 2 left N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 2 mole NH 3 3 mole H 2 1 mole H 2 2.0 g H 2 3.50 g H 2 17.0 g NH 3 1 mole NH 3 = 19.8 g NH 3

10. C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (g) O 2 - limiting reactant, C 3 H 8 - excess reactant What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and 150.0 L of oxygen at STP? 3 = 191 L O 2 5 mole O 2 1 mole C 3 H 8 44.0 g C 3 H 8 75.0 g C 3 H 8 22.4 L O 2 1 mole O 2 3 4 5

11. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) = 58.9 g C 3 H 8 = 90 L CO 2 3 mole CO 2 5 mole O 2 1 mole O 2 22.4 L O 2 150 L O 2 22.4 L CO 2 1 mole CO 2 1 mole C 3 H 8 5 mole O 2 1 mole O 2 22.4 L O 2 150 L O 2 44.0 g C 3 H 8 1 mole C 3 H 8 75.0 g – 58.9 g = 16.1 g C 3 H 8 left

12.  The limiting reactant is completely consumed.  The excess reactant is NOT used up. When solving limiting reactant problems: 1.Balance the chemical equation first. 2.Find the limiting reactant. 3.Use the limiting reactant to determine the moles of the product formed. 4.Calculate the excess from the limiting reactant.