Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting Reactant Reactions don’t always use the correct proportions of reactants.

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Presentation transcript:

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting Reactant Reactions don’t always use the correct proportions of reactants. Many reactions are carried out with an excess amount of one reactant …more than is actually needed. A chemical reaction depends on the reactant that is present in limiting amount - limiting reactant.

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Ham sandwich example 10 slices bread + 5 slices ham 5 ham sandwiches 2 slices of bread + 1 slice ham = 1 sandwich

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Ham sandwich 14 slices bread + 5 slices ham excess reactantlimiting reactant 5 ham sandwiches + 4 slices of bread unreacted

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting reactant example 2Sb (s) + 3I 2 (s)2SbI 3 (s) Determine the limiting reactant and theoretical yield when: moles Sb and 2.40 moles I 2 are mixed g Sb and 2.40g of I 2 are mixed.

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Determing limiting reactant 1. Calculate the amount of product that would be formed if the first reactant were completely consumed. 2. Repeat this calculation for the second reactant. 3. Choose the smaller of the two amounts. This is the theoretical yield. The reactant that produces the smaller amount - limiting reactant.

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting reactant cont… 2Sb (s) + 3I 2 (s)2SbI 3 (s) 1.20mol Sb x 2mol SbI 3 = 1.20mol SbI 3 2mol Sb 2.40mol I 2 x 2mol SbI 3 = 1.60mol SbI 3 3mol I mol SbI 3 is the theoretical yield and Sb is the limiting reactant.

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. Limiting reactant cont… 2Sb (s) + 3I 2 (s)2SbI 3 (s) 1.20g Sb x 1mol Sb x 2mol SbI 3 x 502.5g SbI 3 = 4.95g SbI g Sb 2mol Sb 1mol SbI g I 2 x 1mol I 2 x 2mol SbI 3 x 502.5g SbI 3 = 3.17g SbI g I 2 3mol I 2 1mol SbI g SbI 3 is the theoretical yield and I 2 is the limiting reactant.