2011 topic 01 lecture 3 - limiting reactant and percent yield
Limiting Reactant
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Transcript of Limiting Reactant
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 reactant limiting reactant
5 ham sandwiches + 4 slices of bread
unreacted
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Limiting reactant example
2Sb (s) + 3I2 (s) 2SbI3 (s)
Determine the limiting reactant and theoretical yield when:
1. 1.20 moles Sb and 2.40 moles I2 are mixed.
2. 1.20g Sb and 2.40g of I2 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) + 3I2 (s) 2SbI3 (s)
1.20mol Sb x 2mol SbI3 = 1.20mol SbI3
2mol Sb
2.40mol I2 x 2mol SbI3 = 1.60mol SbI3
3mol I2
1.20 mol SbI3 is the theoretical yield and Sb is the limiting reactant.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Limiting reactant cont…
2Sb (s) + 3I2 (s) 2SbI3 (s)
1.20g Sb x 1mol Sb x 2mol SbI3 x 502.5g SbI3 = 4.95g SbI3
121.8g Sb 2mol Sb 1mol SbI3
2.40g I2 x 1mol I2 x 2mol SbI3 x 502.5g SbI3 = 3.17g SbI3
253.8g I2 3mol I2 1mol SbI3
3.17g SbI3 is the theoretical yield and I2 is the limiting reactant.