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Limiting Reactant

Limiting Reactant. 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.

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Limiting Reactant

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  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. N2(g) + 3 H2(g) 2 NH3(g) H H H H N N N H N H H H N N H H H H 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.

  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. 2 Na (s) + Cl2 (g) 2 NaCl (s) How much NaCl is produced and what is left over when 6.70 mol of Na react with 3.20 mol of Cl2? 6.70 mol Na 1 mol Cl2 = 3.35 mol Cl2 2 mol Na Cl2 - limiting reactant, Na - excess reactant 3.20 mol Cl2 2 mol NaCl = 6.40 mol NaCl 1 mol Cl2

  7. …when 6.70 mol of Na react with 3.20 mol of Cl2? 2 Na (s) + Cl2 (g) 2 NaCl (s) Cl2 - limiting reactant, Na - excess reactant 3.20 mol Cl2 2 mol Na = 6.40 mol Na 1 mol Cl2 6.70 mol - 6.40 mol = 0.30 mol Na excess

  8. How many grams of ammonia can be made from 3.50 g of H2 gas and 18.0 g of nitrogen gas? What’s left? N2(g) + 3 H2 (g) 2 NH3(g) 3.50 g H2 1 mole H2 1 mole N2 28.0 g N2 3 mole H2 2.0 g H2 1 mole N2 = 16.3 g N2 H2 - limiting reactant, N2 - excess reactant

  9. N2(g) + 3 H2 (g) 2 NH3(g) 3.50 g H2 17.0 g NH3 1 mole H2 2 mole NH3 1 mole NH3 3 mole H2 2.0 g H2 = 19.8 g NH3 18.0 g – 16.3 g = 1.68g N2 left

  10. 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 C3H8(g) + O2 (g) CO2 (g) + H2O (g) 5 3 4 75.0 g C3H8 1 mole C3H8 5 mole O2 22.4 L O2 1 mole C3H8 44.0 g C3H8 1 mole O2 = 191 L O2 O2 - limiting reactant, C3H8- excess reactant

  11. C3H8(g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) 150 L O2 1 mole O2 3 mole CO2 22.4 L CO2 5 mole O2 1 mole CO2 22.4 L O2 = 90 L CO2 150 L O2 1 mole O2 1 mole C3H8 44.0 g C3H8 5 mole O2 1 mole C3H8 22.4 L O2 = 58.9 g C3H8 75.0 g – 58.9 g = 16.1 g C3H8 left

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

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