Limiting Reagents
Limiting Reagents
** When a
chemist carries out a reaction, the reactants are usually not present in exact stoichiometric
amounts, that is, in the proportions indicated by the balanced equation.
** Because the
goal of a reaction is to produce the maximum quantity of a useful compound from
the starting materials, frequently a large excess of one reactant is supplied to
ensure that the more expensive reactant is completely converted to the desired product.
**
Consequently, some reactant will be left over at the end of the reaction. The reactant
used up first in a reaction is called the limiting reagent, because the maximum
amount of product formed depends on how much of this reactant was originally present.
When this reactant is used up, no more product can be formed.
** Excess reagents
are the reactants present in quantities greater than necessary to react with the
quantity of the limiting reagent.
The concept of the limiting reagent
** The concept
of the limiting reagent is analogous to the relationship between men and women
in a dance contest at a club.
** If there are
14 men and only 9 women, then only 9 female/male pairs can compete. Five men
will be left without partners.
** The number
of women thus limits the number of men that can dance in the contest, and there
is an excess of men.
Example: synthesis of methanol (CH3OH)
** Consider the
industrial synthesis of methanol (CH3OH) from carbon monoxide and
hydrogen at high temperatures:
** Suppose
initially we have 4 moles of CO and 6 moles of H2 (Figure 1).
** One way to
determine which of the two reactants is the limiting reagent is to calculate
the
number of moles
of CH3OH obtained based on the initial quantities of CO and H2.
** From the
preceding definition, we see that only the limiting reagent will yield the
smaller amount
of the product.
** Starting
with 4 moles of CO, we find the number of moles of CH3OH produced is:
and starting
with 6 moles of H2, the number of moles of CH3OH formed
is:
** Because H2
results in a smaller amount of CH3OH, it must be the limiting
reagent.
Therefore, CO
is the excess reagent.
Note
In
stoichiometric calculations involving limiting reagents, the first step is to decide which
reactant is the limiting reagent. After the limiting reagent has been identified, the
rest of the problem can be solved. the following Example illustrates this
approach.
Solved problems
Urea [(NH2)2CO]
is prepared by reacting ammonia with carbon dioxide:
In one process, 849.2 g of NH3
are treated with 1223 g of CO2.
(a) Which of the two reactants is
the limiting reagent?
(b) Calculate the mass of (NH2)2CO
formed.
(c) How much excess reagent (in
grams) is left at the end of the reaction?
(a) Strategy:
The reactant that forms fewer moles of product
is the limiting reagent because it limits the amount of product that can be
formed. How do we convert from the amount of reactant to the amount of product?
Perform this calculation for each reactant, then compare the moles of product,
(NH2)2CO, formed by the given amounts of NH3 and
CO2 to determine which reactant is the limiting reagent.
Solution:
We carry out two separate calculations. First,
starting with 849.2 g of NH3, we calculate the number of moles of
(NH2)2CO that could be produced if all the NH3
reacted according to the following conversions:
Combining these
steps into one step, we write:
Second, for
1223 g of CO2, the conversions are:
The number of
moles of (NH2)2CO that could be produced if all the CO2
reacted is:
It follows,
therefore, that NH3 must be the limiting reagent because it produces a smaller amount
of (NH2)2CO.
(b) Strategy
We determined the moles of (NH2)2CO
produced in part (a), using NH3 as the limiting reagent. How do we
convert from moles to grams?
Solution:
The molar mass of (NH2)2CO
is 60.06 g. We use this as a conversion factor to convert from moles of (NH2)2CO
to grams of (NH2)2CO:
Check:
Does your
answer seem reasonable? 24.93 moles of product are formed. What is the mass of
1 mole of (NH2)2CO?
(c) Strategy:
Working backward, we can determine the amount
of CO2 that reacted to produce 24.93 moles of (NH2)2CO.
The amount of CO2 left over is the difference between the initial
amount and the amount reacted.
Solution:
Starting with
24.93 moles of (NH2)2CO, we can determine the mass of CO2
that reacted using the mole ratio from the balanced equation and the
molar mass of CO2. The conversion steps are:
The amount of
CO2 remaining (in excess) is the difference between the initial
amount (1223 g) and the amount reacted (1097 g):
Check:
Is this answer reasonable? Can the mass of CO2
remaining be greater than the initial mass of CO2 used in the
reaction?
The last Example
brings out an important point. In practice, chemists usually choose the more
expensive chemical as the limiting reagent so that all or most of it will be
consumed in the reaction. In the synthesis of urea, NH2 is invariably the
limiting reagent because it is much more expensive than CO2
Reference: Organic chemistry / T.W. Graham Solomons , Craig B.Fryhle , Scott A.snyder , / ( eleventh edition) / 2014.
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