Limiting Reactants

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Section 3.7

• Limiting Reactants

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Grilled Cheese Sandwich
Bread Cheese ? Cheese Melt
2 B C ? B2C
100 bread 30 slices ? sandwiches
What is the limiting factor in our ability to
make the maximum about of grilled cheese
sandwiches containing 2 slices of bread and 1
slice of cheese?
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LIMITING REACTANT

• IMPORTANCE
• Calculations of limiting reactant bring
  quantitative understanding to chemical reactions
• These calculations are used in both General and
  Organic Chemistry

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DEFINITIONS

• LIMITING REACTANT
• Completely consumed in a chemical reaction
• Determines the amount of product formed
• The reactant that produces the least amount of
  product

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DEFINITIONS

• THEORETICAL YIELD
• The amount of product that can be made based on
  the amount of the limiting reactant
• ACTUAL YIELD
• The amount of product actually or experimentally
  produced
• THE PERCENT YIELD
• yield (actual/theoretical) x 100

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Limiting Reactants.

• An analogous situation occurs with chemical
  reactions. Consider the reaction
• 2 H2(g) O2(g) ? 2 H2O(l)
• 2 mol 1 mol 2 mol
• If we have exactly 2 mol of H2 and 1 mol of
  O2, then we can make 2 mol of water. But what if
  we have 4 mol of H2 and 1 mol of O2. Now we can
  make only 2 mol H2O with 2 mol H2 left over. In
  this case the O2 is the limiting reagent.
• The limiting reagent is the one with nothing left
  over.

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Container 1
Zumdahl, Zumdahl, DeCoste, World of Chemistry
2002, page 269
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Before and After Reaction 1
Before the reaction
After the reaction
All the hydrogen and nitrogen atoms combine.
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Container 2
Zumdahl, Zumdahl, DeCoste, World of Chemistry
2002, page 270
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Before and After Reaction 2
Before the reaction
After the reaction
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Multiplying an equation through by a common
multiple

• We can multiply all the coefficients in a
  balanced equation by any multiple, and it still
  has the correct ratios of moles. Thus, if we
  have
• Zn(s) 2HCl(aq) ? ZnCl2(aq) H2(g)
• 1 mole 2 moles 1 mole 1
  mole
• If we have 2 moles of Zn(s), this gives (x 2)
• 2 moles 4 moles 2 moles
  2 moles
• or if we have 0.5 moles Zn(s) we have (x 0.5)
• 0.5 moles 1 mole 0.5 moles
  0.5 moles

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METHODS USED TO DETERMINE THE LIMITING REACTANT

• Calculate the moles needed of each reactant and
  compare with the moles given
• II. Divide the moles of each reactant by its
  stoichiometric coefficient and then compare them
• Calculate the moles of product produced by each
  reactant and compare them

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• Example I. Consider the reaction of H2 and N2 to
  give NH3, and assume we have 3.0 mol N2 and 6.0
  mol H2.
• We have the balanced equation
• N2(g) 3 H2(g) ? 2 NH3(g)
• 1 mol 3 mol 2 mol
• Factor moles N2 we have
• moles N2 in equation
• 3.0 mol N2
• 1.0 mol N2
• 3.0 (multiply all coefficients in
• balanced equation by
  this factor)

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• Multiply all coefficients by factor (x 3)
• N2(g) 3 H2(g) ? 2 NH3(g)
• 1 mol 3 mol 2 mol
• 3 mol 3 x 3 9 mol 3 x 2 6 mol
• Try N2 as limiting reagent
• 3 mol N2 requires how many moles H2?
• 3 x 3 9 mol
• We only have 6 mol H2, so H2 is the
• limiting reagent.

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Example II. Divide the moles of each reactant by
its stoichiometric coefficient

• Consider the following reaction
• 2 Na3PO4(aq) 3 Ba(NO3)2(aq) ?Ba3(PO4)2 6
  NaNO3
• How much Ba3(PO4)2 can be formed if we have in
  the solutions 3.50 g sodium phosphate and 6.40 g
  barium nitrate?

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Step 1. Convert to moles

• First work out numbers of Moles
• Na3PO4 3.50 g x 1 mol 0.0213 mol
• 164 g
• Ba(NO3)2 6.40 g x 1 mol 0.0245 mol
• 261 g

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Step 2. Divide moles by its stoichiometric
coefficient

• 2 Na3PO4(aq) 3 Ba(NO3)2(aq) ?Ba3(PO4)2 6
  NaNO3
• Na3PO4 0.0213 mol 0.01065
• 2 mol
• Ba(NO3)2 0.0245 mol 0.00817 LR
• 3 mol

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• Example III. Calculate the amount of product
  produced by each reactant
• 1N2(g) 3H2(g) ? 2NH3(g)
• Given 3.0 mole 6.0 mole
• 3.0 mol N2 x 2 mol NH3 6.0 mol NH3
• 1 mol N2
• 6.0 mol H2 x 2 mol NH3 4.0 mol NH3
  (theoretical yield)
• 3 mol H2
• The reactant that produces the least amount of
  product is the L.R.?H2

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Practice Exercise

• Zn metal (2.00 g) plus solution of AgNO3 (2.50
  g) reacts according to
• Zn(s) 2 AgNO3(aq) ? Zn(NO3)2 2 Ag(s)
• 1 mol 2 mol
• Which is the limiting reagent?
• How much Zn will be left over?

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Step 1. Convert to moles

• Zn 65 g/mol
• AgNO3 108 14 (3 x 16) 170 g/mol
• Zn 2.0 g x 1 mol 0.0308 mol
• 65 g
• AgNO3 2.50 g x 1 mol 0.0147 mol
  170 g

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Step 2. Guess limiting reagent

• Zn(s) 2 AgNO3(aq) ? Zn(NO3)2 2 Ag(s)
• 1 mol 2 mol
• 0.0308 0.0147
• In this case it seems clear that AgNO3 must be
  the limiting reagent, because the equation says
  we must have 2 mols of AgNO3 for each mol of
  Zn(s), but in fact we have more moles of Zn(s).

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• We can check this by dividing the moles of each
  reactant by their coefficients
• AgNO3 0.0147/2 0.00735
• Zn 0.0308/1 0.0308
• Zn(s) 2 AgNO3(aq) ? Zn(NO3)2 2 Ag(s)
• 1 mol 2 mol
• 0.0308 0.00735
• We in fact have 0.0305 mol of Zn, which is more
  than the 0.00735 mol of AgNO3, so AgNO3 is
  clearly the limiting reactant.

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How much Zn is left over?

• Use the limiting reactant to determine this
• 0.0147 mol AgNO3 x 1 mol Zn x 65 g Zn
• 2 mol AgNO3
  1 mol Zn
• 0.47775 g Zn
• Subtract this from the amount of Zn available
• 2.00 g Zn - 0.4775g Zn 1.52 g Zn in excess

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Homework

• 3.71-3.74 on pages 115-116

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Percent Yield

• Theoretical yields
• The quantity of product that forms if all of the
  limiting reagent reacts is called the theoretical
  yield. Usually, we obtain less than this, which
  is known as the actual yield.
• Percent yield actual yield x 100
• Theoretical yield

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Problem

• 10.4 g of Ba(OH)2 was reacted with an excess of
  Na2SO4 to give a precipitate of BaSO4. If the
  reaction actually yielded 11.2 g of BaSO4, what
  is a) the theoretical yield of BaSO4 and b) what
  is the percentage yield of BaSO4?
• The balanced equation for the reaction is
• Ba(OH)2(aq) Na2SO4(aq) ? BaSO4(s) 2 NaOH(aq)

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Step 1. Convert to moles

• Ba(OH)2(aq) Na2SO4(aq) ? BaSO4(s) 2 NaOH(aq)
• 1 mole 1 mole 1 mole 2 moles
• Moles Ba(OH)2
• Mol. Mass Ba(OH)2 137.3 2 x (16.0 1.0)
• 171.3 g/mol
• Moles 10.4 g x 1 mol 0.0607 moles
• 171.3 g

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Step 2. Work out how much BaSO4 will be formed

• Ba(OH)2(aq) Na2SO4(aq) ? BaSO4(s) 2 NaOH(aq)
• 1 mole 1 mole 1 mole 2 moles
• 0.0607 moles 0.0607 moles
• When it says that one reagent is in excess, that
  means we do not have to worry about that reagent,
  and the other one is the limiting reagent, in
  this case the BaSO4.
• We see that 1 mole of Ba(OH)2 will produce 1
  mole of BaSO4. Our factor is thus 0.0607, and we
  will get 0.0607 moles of BaSO4.

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Convert actual yield to percentage yield

• Percent yield actual yield x 100
• Theoretical yield
• 11.2 g x 100
• 14.29 g
• 78.4 yield