LIMITING REAGENT

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LIMITING REAGENT
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Just a Quick Note

• Gas occupies 22.4L/mol at STP (0oC)
• Gas occupies 24.5L/mol at NTP (25oC)

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Limiting Reactant
In a chemical reaction where arbitrary amounts of
reactants are mixed and allowed to react, the one
that is used up first is the limiting reactant.
A portion of the other reactants remains.
There is a systematic procedure for finding the
limiting reagent based on the reactant ratio (RR)
defined as the ratio of the number of moles of a
reactant to its coefficient in a balanced
chemical equation. The reagent with the smallest
reactant ratio is the limiting reactant.
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The Limiting Reagent

• In most chemical reactions, not all of the
  reactants are used up because they are not
  present in the exact proportions required of the
  reaction
• Eg. If Pb(NO3)2 CuSO4 ? PbSO4 Cu(NO3)2
• you would need exactly 313g (1mol) of Pb(NO3)2
  and
• 303g (1mol) of CuSO4 to fully use each of the
  reactants
• The limiting reagent is the reactant which is
  used up first in a chemical reaction (i.e. it
  being used up, stops the rest of the reactants
  from being able to fully react)
• The limiting reagent can be determined
  mathematically using the Mole

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Balanced reaction! Defines stoichiometric
ratios!
Unbalanced (i.e., non-stoichiometric)
mixture!
Limited by syrup!
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For a Reaction of the Form aA bB cC
dD If compounds A and B are present in the mole
amounts called for in the balanced reaction, then
the following equation is valid
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To proceed, first calculate the reactant ratios
for all of the reactants
aA bB cC dD
From among these, choose (RR)min, the smallest
reactant ratio. This identifies the limiting
reactant.
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Limiting Reactant EXAMPLE

• 2Al(s) 6HCl(g) 2AlCl3(s)
  3H2(g)
• Consider the reaction above. If we react 30.0 g
  Al and 20.0 g HCl, how many moles of aluminum
  chloride will be formed?
• 30.0 g Al 20.0g HCl
• Limiting reactant reactant with RRmin

MOLES M/MM 30 / 27 1.1111 RRa (Moles
of Al) / a 1.1111 / 2 0.5555
MOLES M/MM 20/ (1 35.5) 0.5479 RRb
(Moles of HCl) / b 0.5479 / 6
0.09132
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Finding amount of Product from Limiting Reagent
2Al(s) 6HCl(g) 2AlCl3(s)
3H2(g)
Calculate the Moles of AlCl3 from Moles of
Limiting Reagent x (need/have) 0.5479 x 2/
6 0.18264

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• 4KO2 2H2O ? 4KOH 3O2 0.15mol KO2 and 0.10mol
  H2O
• Which is the limiting reagent?
• How many moles of O2 can be produced?
• What is the volume of O2 is produced at STP?
• CO H2 ? CH3OH 35.4g CO and 10.2g H2
• Balance the equation
• Which is the limiting reagent?
• How many grams of the excess reagent are left at
  the end of reaction?
• ( Hint Need / have x sentence) use this to
  calculate the moles of H2 used

• .
• KO2 is the limiting reagent
• 0.11mol of O2
• 2.52 L
• .
• CO 2H2 ? CH3OH
• CO is the limiting reagent
• 5.1g of H2

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