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Limiting Reactant

Limiting Reactant. 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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Limiting Reactant

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  1. Limiting Reactant

  2. Identify the limiting reactantand 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. 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- determinesthe 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: Record what you HAVE Step 2: Calculatewhat you NEED Pick one reactant and calculate how much of the other you will need. Step 3: Identify the limiting reactant Step 4: Use limiting reactant to determine the amount of product.

  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 3.20 mol HAVE NEED Pick one reactant and calculate the other 6.70 mol Na 1 mol Cl2 = 3.35 mol Cl2 (need) 2 mol Na

  7. 2 Na (s) + Cl2 (g) 2 NaCl (s) HAVE 6.70 mol 3.20 mol NEED 3.35 mol Pick one reactant and calculate the other 3.20 mol Cl2 2 mol Na = 6.40 mol Na(need) 1 mol Cl2 Both calculations lead to the same conclusion: Have too much Na and Don’t have enough Cl2

  8. 2 Na (s) Cl2 (g) 2 NaCl(s) HAVE 6.70 mol 3.20 mol - NEED 6.40 mol 3.35 mol Na - excess reactant Cl2- limiting reactant 3.20 mol Cl2 2 mol NaCl = 6.40 mol NaCl 1 mol Cl2 You could use your data to calculate exactly how much excess is left over: 6.70 mol - 6.40 mol = 0.30 mol Na excess

  9. 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) HAVE 18.0 g 3.50 g 3.90 g NEED 16.3 g 3.50 g H2 1 mol H2 1 mol N2 28.0 g N2 = 16.3 g N2 3 mol H2 2.0 g H2 1 mole N2 18.0 g N2 1 mol N2 3 mol H2 2.02 g H2 = 3.90 g H2 1 mol N2 28.0 g N2 1 mole H2

  10. N2(g) + 3 H2 (g) 2 NH3(g) HAVE 18.0 g 3.50 g 3.90 g NEED 16.3 g N2 - excess reactant H2- limiting reactant 3.50 g H2 17.0 g NH3 1 mol H2 2 mol NH3 1 mol NH3 3 mol H2 2.0 g H2 = 19.8 g NH3 18.0 g – 16.3 g = 1.70g N2 left

  11. What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and 150.0 L of oxygen at STP? C3H8(g) + O2 (g) CO2 (g) + H2O (g) 5 3 4 HAVE 75.0 g 150.0 L 191 L NEED 58.9 g 75 g C3H8 22.4 L O2 1 mol C3H8 5 mol O2 = 191 L O2 1 mol O2 1 mol C3H8 44 g C3H8 150 L O2 44 g C3H8 1 mol O2 1 mol C3H8 = 58.9 g C3H8 5 mol O2 22.4 L O2 1 mol C3H8

  12. C3H8 + 5 O2 3 CO2 + 4 H2O HAVE 75.0 g 150.0 L 191 L NEED 58.9 g C3H8- excess reactant O2 - limiting reactant 150 L O2 1 mol O2 3 mol CO2 22.4 L CO2 = 90 L CO2 5 mol O2 1 mol CO2 22.4 L O2 75.0 g – 58.9 g = 16.1 g C3H8 left

  13. 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 limiting reactant to determine the product • Calculate the excess

  14. CAN YOU / HAVE YOU? • Identifythe limiting reactantand 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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