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Limiting Reactants

Continuing Stoichiometry …. Limiting Reactants. The idea. In every chemical reaction, there is one reactant that will be run out (called the limiting reactant ) . This will limit the amount of product that can form. The reaction will stop at that point.

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Limiting Reactants

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  1. Continuing Stoichiometry… Limiting Reactants

  2. The idea. • In every chemical reaction, there is one reactant that will be run out (called the limiting reactant) . • This will limit the amount of product that can form. • The reaction will stop at that point. • There are then also excess reactants.

  3. Sample Problem • S8 + 4 Cl2  4 S2Cl2 If you have 200.0 g S8 and 100.0 g Cl2, what is the limiting reactant?

  4. Step 1 • First, find the moles of each reactant. • Amount of each reactant will be given • Usually it will be in grams. This is a gram to mole conversion.

  5. The Answers 200.0 g S8 1 mole S8 x = .78 mol S8 1 256 g S8 100.0 g Cl2 1 mole Cl2 x = 1.41 mol Cl2 1 71 g Cl2

  6. If moles are Given… • If moles are given, problem starts here…

  7. Step 2: Compare Mole Ratios • A) What is the mole to mole ratio from the balanced chemical equation? • S8 + 4 Cl2  4 S2Cl2 Moles S8 = 1 Moles Cl2 = 4 Need 4 Cl2 for each S8!

  8. Step 2 cont’d • B) What is the mole to mole ratio calculated from the starting conditions? Moles S8 = .78 mol S8 Moles Cl2 = 1.41 mol Cl2 1.41 mol Cl2 = 1.81 mol Cl2 for 1 mol S8 .78 mol S8

  9. Is this enough? • NO! • Therefore, Cl2 is the limiting reactant Now practice some LR calculations!

  10. Calculating the Amount of Product Formed • Complete a mole to gram calculation using the Limiting Reactant • In our example: • LR: Cl2 = 1.41 moles Try it!

  11. The Answer • S8 + 4 Cl2  4 S2Cl2 1.41 mol Cl2 4 mol S2Cl2 135 g S2Cl2 x x 1 4 mol Cl2 1 mol S2Cl2 = 190.35 g S2Cl2

  12. Analyzing the Excess Reactant • You care about two things: • How much of the excess reactant reacted • How much of the excess reactant remains

  13. Excess that Reacted • Complete a mole to gram calculation using the limited and excess reactants • In our example: • Limiting Reactant: Cl2 = 1.41 mol • Excess Reactant: S8

  14. The Calculation • S8 + 4 Cl2  4 S2Cl2 1.41 mol Cl2 1 mol S8 256 g S8 x x 1 4 mol Cl2 1 mol S8 = 90.24 g S8

  15. Excess Remaining • Subtract amount that reacted from the amount that you started with • In example: • Amount Reacted = 90.24 g S8 • Amount Starting With = 200.0 g S8

  16. The Answer 200.0 g - 90.24 g = 109.76 g S8 in excess

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