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Limiting Reactant. Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data. Include: theoretical.

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Presentation on theme: "Limiting Reactant. Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data. Include: theoretical."— Presentation transcript:

1 Limiting Reactant

2 Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data. Include: theoretical yield, experimental yield Additional KEY Terms Excess reactant

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: Record what you HAVE Step 2: Calculate what you NEED Pick one reactant and calculate how much of the other you will need. Step 3: Determine the limiting reactant and excess reactant Step 4: Use limiting reactant to determine the amount of product.

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) 1 mol Cl 2 2 mol Na 6.70 mol Na Cl 2 (need) Pick one reactant and calculate the other HAVE NEED 6.70 mol3.20 mol = 3.35 mol

7 2 Na (s) + Cl 2 (g) 2 NaCl (s) 1 mol Cl 2 2 mol Na 3.20 mol Cl 2 Both calculations lead to the same conclusion: Cl 2 - limiting reactant Na - excess reactant Pick one reactant and calculate the other HAVE 6.70 mol 3.20 mol NEED 3.35 mol Na (need) = 6.40 mol

8 = 6.40 mol NaCl 1 mol Cl 2 2 mol NaCl 3.20 mol Cl 2 You could use your data to calculate exactly how much excess is left over: 6.70 mol mol = 0.30 mol Na excess 2 Na (s) + Cl 2 (g) 2 NaCl (s) Cl 2 - limiting reactant Na - excess reactant HAVE 6.70 mol 3.20 mol NEED 6.40 mol 3.35 mol

9 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 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 mol N 2 3 mol H 2 1 mol H g H g H g N 2 1 mole N 2 = 16.3 g N 2 HAVE 18.0 g 3.50 g NEED16.3 g 3 mol H 2 1 mol N g N g N g H 2 1 mole H 2 = 3.90 g H g

10 18.0 g – 16.3 g = 1.70g N 2 left N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 2 mol NH 3 3 mol H 2 1 mol H g H g H g NH 3 1 mol NH 3 = 19.8 g NH 3 H 2 - limiting reactant N 2 - excess reactant HAVE 18.0 g 3.50 g NEED16.3 g 3.90 g

11 What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and L of oxygen at STP? HAVE 75.0 g L NEED58.9 g 191 L C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (g) = 191 L O 2 5 mol O 2 1 mol C 3 H 8 44 g C 3 H 8 75 g C 3 H L O 2 1 mol O 2 = 58.9 g C 3 H 8 1 mol C 3 H 8 5 mol O 2 1 mol O L O L O 2 44 g C 3 H 8 1 mol C 3 H 8

12 75.0 g – 58.9 g = 16.1 g C 3 H 8 left O 2 - limiting reactant C 3 H 8 - excess reactant HAVE 75.0 g L NEED58.9 g 191 L C 3 H O 2 3 CO H 2 O = 90 L CO 2 3 mol CO 2 5 mol O 2 1 mol O L O L O L CO 2 1 mol CO 2

13  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 limiting reactant to determine the product 4.Calculate the excess

14 CAN YOU / HAVE YOU? Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data. Include: theoretical yield, experimental yield Additional KEY Terms Excess reactant


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