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Limiting Reactions

Chemistry. Limiting Reactions. Limiting Reactions. To drive a car, you need fuel and oxygen. When the fuel runs out, you can’t go any further even though there is still plenty of oxygen! (What is the limiting reactant?). The Limiting Reactant Forms the Least Amount of Product.

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Limiting Reactions

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  1. Chemistry Limiting Reactions

  2. Limiting Reactions • To drive a car, you need fuel and oxygen. When the fuel runs out, you can’t go any further even though there is still plenty of oxygen! (What is the limiting reactant?)

  3. The Limiting Reactant Forms the Least Amount of Product • The limiting reactant is the substance that controls the amount of the product that can be formed from the reaction. • The excess reactant is the substance that is not used up completely in a reaction.

  4. What Substance is the Limiting Reactant in this Reaction?

  5. What is the Limiting Reactant in this Reaction?

  6. Limiting Reactant Problem • If you have two GIVENs, it MUST be a limiting reactant problem. • Solve the stoichiometry problem twice—once for each GIVEN. • The correct answer is the SMALLEST answer—Why? Because the reaction will stop once one of the reactants is gone.

  7. Example Problem • How many grams of Cu can be formed when 167.4 g of Fe reacts with 399.2 g of CuCl2? • 2 Fe + 3 CuCl2 → 3 Cu + 2FeCl3 • Solve the Stoichiometry problem twice, once for Fe and once for CuCl2.

  8. Example Problem • How many grams of Cu can be formed when 167.4 g of Fe reacts with 399.2 g of CuCl2? • 2 Fe + 3 CuCl2 → 3 Cu + 2FeCl3 • 167.4g Fe 1 mol Fe 3 mol Cu 63.5 g Cu 284.8g Cu • 55.8 g Fe 2 mol Fe 1 mol Cu • 399.2g CuCl2 1 mol 3 mol Cu 63.5 g Cu 188.5 g Cu • 134.5 3 mol 1 mol Cu • CuCl2CuCl2 This is the smallest! It is the answer!

  9. Excess Reactant • To find how much excess you have, you must take the limiting reactant (Solve for that first!), then solve for the excess reactant.

  10. Example Problem for Excess Reactant • In the previous problem, you determined that CuCl2 was the limiting reactant. • How did you determine this??? • Because the amount of Cu formed by Fe and CuCl2 was limited by the CuCl2. • It was only 188.5 g Cu from CuCl2 (compared to 285.8 g from the Fe) • The Limiting Reactant is CuCl2; The Excess Reactant is Fe.

  11. Example Problem for Excess Reactant: • How many grams of excess Fe is left over from the previous problem? • NOTE: Since CuCl2 was the limiting reactant, the amount of CuCl2 becomes the GIVEN: • 399.2 g CuCl2 1 mol 2 mol Fe 55.8g Fe 110.4g Fe • 134.5 g 3 mol 1 mol Fe needed • CuCl2 CuCl2 • We AREN’T Done Yet!!!!

  12. Example Problem: Excess Reactant • The stoichiometry problem shows that you needed 110.4 g of Fe. Since you started with 167.4 g of Fe, all you need to do is subtract the two to find out how much excess you have. • 167.4 g Fe (Given in original problem, step 1) • -110.4 g Fe (Needed) • =57.0 g Fe (excess)

  13. Percent Yield: • Finding % Yield: Actual Yield X 100= % yield • Theoretical Yield

  14. Actual Yield • Actual Yield Must be found experimentally. You must do the experiment yourself or you will be given the information. You will see words like: produced and formed in a problem that is giving you the actual yield.

  15. Theoretical Yield • Theoretical Yield is done using stoichiometry. You would take the given and find out how much of the product should be formed. • In our problem, the yield of 188.5 g Cu is the Theoretical Yield: • 399.2g CuCl2 1 mol 3 mol Cu 63.5 g Cu 188.5 g Cu • 134.5 3 mol 1 molCu ↑ • CuCl2CuCl2 This is the Theoretical Yield because it is the smallest

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