A chemical change may be defined as: "A process in which a new
substance is formed"
Syllabus reference R2.1.3
Reactivity 2.1.3 - The limiting reactant determines the theoretical yield.
- Identify the limiting and excess reactants from given data.
Guidance
- Distinguish between the theoretical yield and the experimental yield.
Tools and links
- Tool 1, Inquiry 1, 2, 3 - What errors may cause the experimental yield to be i) higher and ii) lower than the theoretical yield?
Excess reagent
When there is more of one of the reactants present than the required amount,
the extra will not have anything to react with. This should be apparent bearing
in mind the particulate nature of matter. If two molecules of hydrogen react
with one molecule of oxygen to make two molecules of water, then any 'extra'
molecules of hydrogen (or oxygen for that matter) will be unable to react.
Three hydrogen molecules trying to react with one oxygen molecule
Here you can see that the extra hydrogen molecule (at the lower right of
the diagram) remains unreacted. We call this the excess reagent.
An extra amount of chemical over and above that which is needed for complete
reaction is called the excess. A reagent in excess (i.e. one of which there
is more than the amount needed) cannot completely react. Some of it can react,
but the rest simply remains unreacted after the reaction has finished.
Calculating the excess
To find the excess reagent, the first stage is to calculate the number of
moles of each reagent in the reaction. Then the stoichiometry of the equation
shows the relative number of moles reacting in an ideal situation. The excess
is found by substituting the number of moles of the first reagent (reacting
chemical) in the given situation and seeing how many moles of the second reagent
is required for complete reaction.
If there is more than enough of the second reagent then it is in excess.
If there is not enough of the second reagent then the first reagent is in
excess.
Example: 64 grams
of sulfur react with 64 grams of iron
Fe + S →
FeS
1 mole of iron reacts completely with 1 mole of sulfur.
RAM Fe = 56, RAM S=32.
From the quantities given we have 64/32 moles of sulfur and 64/56 moles
of iron
i.e. 2 moles of sulfur and 1.143 moles of iron
But as they always react in a 1 : 1 mole ratio then 2 moles of sulfur
would need 2 moles of iron. Clearly there is not enough iron to react
with all the sulfur. Therefore some of the sulfur must remain unreacted
once all of the iron is used up.
The sulfur is said to be in excess.
Only 1.143 moles of the sulfur can react with 1.143 moles of iron.
This means that 2 - 1.143 moles of sulfur is left unreacted = 0.857
moles
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Limiting reagent
In the above example it can be seen that the component that determines the
amount of sulfur that can react is the iron. All of the iron manages to react
and it is this that determines the quantity of products formed. The iron is
said to LIMIT the reaction. It is called the limiting reagent.
To determine the limiting reagent (and to find out which of the reactants
is in excess) the stoichiometry of the reaction must be considered.
Procedure
- Firstly find the relative number of moles of each component in the balanced
equation.
- Then convert the data given in the question under study into moles.
- Now by inspection see which one of the components will completely react
and which one will be in excess.
Example: 25cm3
of 0.2 M sodium hydroxide solution is mixed with 80 cm3 of
0.05 M sulfuric acid.
2NaOH + H2SO4 →
Na2SO4 + 2H2O
From the equation 2 moles of sodium hydroxide react completely with
1 mole of sulfuric acid
From the data given:
25cm3 of 0.2 M sodium hydroxide = 0.025 x 0.2 moles = 0.005
moles
80 cm3 of 0.05 M sulfuric acid = 0.08 x 0.05 = 0.004 moles
From the equation mole ratios, 0.005 moles of sodium hydroxide would
react completely with 0.005/2 = 0.0025 moles of sulfuric acid. But there
are 0.004 moles of sulfuric acid (i.e. more than enough). Some of the
sulfuric acid will remain unreacted - it is in EXCESS.
The LIMITING reagent is the sodium hydroxide as it is this alone that
determines the amount of product formed
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Worked examples - limiting reagent
Find the limiting reagent in each of the following questions
Q361-01 15 g of calcium metal
burns in 11.35 dm3 of oxygen (measured at STP, 1 mole of gas occupies
22.7 dm3)
Answer
|
Equation for the reaction:
2Ca + O2 →
2CaO
mole ratio from the equation: 2 moles of calcium reacts completely
with 1 mole of oxygen gas.
Data given [RAM: Ca = 40]
Moles of calcium = 15/40 = 0.375 moles
Moles of oxygen = 11.35/22.7 = 0.5 moles
From the equation: 0.375 moles of calcium would require 0.375/2 moles
of oxygen = 0.1875 moles
There is more than enough oxygen so it is in EXCESS
The LIMITING reactant is the calcium
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Q361-02 12.35 g of copper
carbonate dissolves in 500 cm3 of 0.1M HCl
Answer
|
Equation for the reaction:
CuCO3 + 2HCl →
CuCl2 + H2O + CO2
mole ratio from the equation: 1 moles of copper carbonate reacts
completely with 2 moles of HCl.
Data given [Mr : CuCO3 = 63.5 + 12 + (3 x 16)
= 123.5]
Moles of CuCO3 = 12.35/123.5 = 0.1 moles
Moles of HCl = 0.5 x 0.1 = 0.05 moles
From the equation: 0.1 moles of CuCO3 would require 0.2
moles of HCl
There is not enough HCl so it is the LIMITING reactant.
The copper carbonate is in excess.
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Q361-03 1.2 g of magnesium
is dissolved in 50 cm3 of 2M sulfuric acid.
Answer
|
Equation for the reaction:
Mg + H2SO4 →
MgSO4 + H2
Ar of magnesium = 24, therefore 1.2 g represents 1.2/24
moles = 0.05 moles
From the equation, 1 mole of magnesium requires 1 mole of sulfuric
acid
Therefore 0.1 moles magnesium requires 0.1 moles of sulfuric acid.
50 cm3 of 2M sulfuric acid is equivalent to 0.05 x 2 =
0.1 moles of sulfuric acid.
There is more than enough sulfuric acid - it is in EXCESS
The limiting reagent is the magnesium
|
Q361-04 7.95 g of copper(II)
oxide is dissolved in 50 cm3 of 2M hydrochloric acid.
Answer
|
Equation for the reaction:
CuO + 2HCl →
CuCl2 + H2O
Mr of copper(II) oxide = 63.5 + 16 = 79.5
7.95 g represents 7.95/79.5 moles = 0.1 moles
From the equation, 1 mole of copper oxide requires 2 moles of hydrochloric
acid
Therefore 0.1 moles copper oxide requires 0.2 moles of hydrochloric
acid.
50 cm3 of 2M hydrochloric acid is equivalent to 0.05 x
2 = 0.1 moles of hydrochloric acid.
There is not enough hydrochloric acid to react with all the copper(II)
oxide - it is the LIMITING REAGENT
The copper oxide is in EXCESS.
|
Q361-05 1.27 g of iodine reacts
with 2.7g aluminium metal [I=127, Al=27]
Answer
|
Equation for the reaction:
2Al + 3I2 →
2AlI3
2 moles of aluminium require 3 moles of iodine.
2.7 g aluminium is equal to 2.7/27 moles aluminium = 0.1 moles
this would require 0.1 x 3/2 = 0.15 moles of iodine
1.27 g iodine = 1.27/254 moles = 0.005 moles iodine.
There is not enough iodine to react with all of the aluminium. The
iodine is the LIMITING REAGENT.
The aluminium is in EXCESS.
|
Q361-06 2.27 dm3
of chlorine gas is absorbed by 100cm3 of 2M sodium hydroxide (measured
at STP, 1 mole of gas occupies 22.7 dm3)
Answer
|
Equation for the reaction
NaOH + Cl2 →
NaOCl + HCl
one mole of chlorine requires 1 mole of sodium hydroxide
Moles of chlorine = 2.27/22.7 = 0.1 moles
This would require 0.1 moles NaOH
Moles of NaOH = 0.1 x 2 = 0.2 moles
There is more than enough NaOH - it is in EXCESS.
The chlorine is the LIMITING REAGENT.
|
Q361-07 11.35 dm3
of ammonia gas is absorbed by 1dm3 of 2M sulfuric acid (measured
at STP, 1 mole of gas occupies 22.7 dm3)
Answer
|
Equation for the reaction
2NH3 + H2SO4 →
(NH4)2SO4
2 moles of ammonia require 1 mole of sulfuric acid
Moles of ammonia = 11.35/22.7 = 0.5 moles
This would require 0.5/2 moles of sulfuric acid = 0.25 moles
Moles of sulfuric acid = 1 x 2 = 2 moles
There is more than enough sulfuric acid - it is in EXCESS.
The ammonia is the LIMITING REAGENT.
|
Q361-08 45.4 dm3
of carbon dioxide gas is absorbed by 1dm3 of 2M sodium hydroxide
(measured at STP, 1 mole of gas occupies 22.7 dm3)
Answer
|
Equation for the reaction
CO2 + 2NaOH →
Na2CO3 + H2O
1 mole of carbon dioxide requires 2 moles of sodium hydroxide
Moles of carbon dioxide = 45.4/22.7 = 2 moles
This would require 2 x 2 moles of sodium hydroxide = 4 moles
Moles of sodium hydroxide = 1 x 2 = 2 moles
There is not enough sodium hydroxide - it is the LIMITING REAGENT
The carbon dioxide is in EXCESS..
|
Q361-09 1.27 g of iodine [I=127]
reacts with 50cm3 of 0.5M sodium thiosulfate according to the equation:
I2 + 2Na2S2O3 →
Na2S4O6 + 2NaI
Answer
|
Mr of I2 = 2 x 127 = 254
Moles of iodine present = 1.27/254 = 0.005 moles
From the equation 1 mole of iodine requires 2 moles of sodium thiosulfate
Therefore 0.005 moles of iodine would require 0.005 x 2 = 0.01 moles
of sodium thiosulfate
Moles of sodium thiosulfate = 0.05 x 0.5 moles = 0.025 moles
This is more than enough sodium thiosulfate to react with all of
the iodine. The sodium thiosulfate is in EXCESS.
The iodine is the LIMITING REAGENT
|
Q361-10 In a thermite reaction
5.4 g of aluminium reduces 12.0 g of iron(III) oxide. Calculate the limiting
reagent and the mass of iron produced. [Ar: Fe=56, Al=27, O=16]
Answer
|
Equation for the reaction:
2Al + Fe2O3 →
Al2O3 + 2Fe
2 moles of aluminium require 1 mole of iron(III) oxide.
From the data given
Moles of aluminium = 5.4/27 = 0.2 moles
This would require 0.2/2 = 0.1 moles of iron(III) oxide
Mr:iron(III) oxide = (56 x 2) + (16 x 3) = 160
12.0 g of iron(III) oxide = 12.0/160 = 0.075 moles
This is not enough to react with all of the aluminium therefore the
iron(III) oxide is the LIMITING REAGENT.
The limiting reagent determines the amoiunt of final product. There
is 0.075 moles of iron(III) oxide and from the equation we see that
this will produce twice as many moles of iron. Therefore moles of
iron produced = 2 x 0.075 = 0.15 moles.
Mass of iron produced = moles x 56
Mass of iron produced = 0.15 x 56 = 8.4
g
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Now test yourself
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