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Limiting Reactant.  Determine which reactant is left over in a reaction.  Identify the limiting reactant and calculate the mass of the product.  Calculate.

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Presentation on theme: "Limiting Reactant.  Determine which reactant is left over in a reaction.  Identify the limiting reactant and calculate the mass of the product.  Calculate."— Presentation transcript:

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 mol 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 g H g H 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 g H g H 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 L of oxygen at STP? 3 = 191 L O 2 5 mole O 2 1 mole C 3 H g C 3 H g C 3 H L O 2 1 mole O

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 L O L O L CO 2 1 mole CO 2 1 mole C 3 H 8 5 mole O 2 1 mole O L O L O g C 3 H 8 1 mole C 3 H 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.


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