
DDT, an insecticide harmful to fish, birds and humans is produced by the following reaction:
$2{{C}_{6}}{{H}_{5}}Cl\left( chlorobenzene \right)+{{C}_{2}}HOC{{l}_{3}}\left( chloral \right)\to {{C}_{14}}{{H}_{9}}C{{l}_{5}}\left( DDT \right)+{{H}_{2}}O$
In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral. What mass of the excess reactant is left over?
Answer
529.5k+ views
Hint:The stoichiometry of the given reaction will help solve this problem more efficiently as, stoichiometry and the general mole concepts will give the answer.
The given problem is based on DDT and generalised information about that; but in actual practise the illustration can be solved by mole concepts only.
Complete step-by-step answer:Let us discuss some key points and then solve the actual problem;
Limiting reactant-
In a chemical reaction, if the reactant is totally consumed when the reaction ends then that reactant is known as limiting reactant.
Excess reactant-
The reactant that is present in higher amounts than actually required for the reaction is known as excess reactant.
Illustration-
Given that,
$2{{C}_{6}}{{H}_{5}}Cl\left( chlorobenzene \right)+{{C}_{2}}HOC{{l}_{3}}\left( chloral \right)\to {{C}_{14}}{{H}_{9}}C{{l}_{5}}\left( DDT \right)+{{H}_{2}}O$
Mass of chlorobenzene = 1142 g
Mass of chloral = 485 g
Molar mass of chlorobenzene = 112.5 g/mol
Molar mass of chloral = 147.5 g/mol
Now,
According to the stoichiometry,
We require 2 moles of chlorobenzene and a mole of chloral to form a mole of DDT. Thus, chlorobenzene is the excess reactant.
So, for given masses;
No. of moles of chlorobenzene $=\dfrac{1142g}{112.5g/mol}=10.15mol$
No. of moles of chloral $=\dfrac{485g}{147.5g/mol}=3.29mol$
No. of moles of excess reactant i.s. left reactant after the reaction $=10.15mol-\left( 2\times 3.29mol \right)=3.57mol$
Thus, mass of excess reactant $=3.57mol\times 112.5g/mol=401.63g$
Note:Do note to use formulas and the units properly as some confusion can arise while solving the problem.
Before solving such problems, do note that you must be aware of the concepts of limiting and excess reactants.
The given problem is based on DDT and generalised information about that; but in actual practise the illustration can be solved by mole concepts only.
Complete step-by-step answer:Let us discuss some key points and then solve the actual problem;
Limiting reactant-
In a chemical reaction, if the reactant is totally consumed when the reaction ends then that reactant is known as limiting reactant.
Excess reactant-
The reactant that is present in higher amounts than actually required for the reaction is known as excess reactant.
Illustration-
Given that,
$2{{C}_{6}}{{H}_{5}}Cl\left( chlorobenzene \right)+{{C}_{2}}HOC{{l}_{3}}\left( chloral \right)\to {{C}_{14}}{{H}_{9}}C{{l}_{5}}\left( DDT \right)+{{H}_{2}}O$
Mass of chlorobenzene = 1142 g
Mass of chloral = 485 g
Molar mass of chlorobenzene = 112.5 g/mol
Molar mass of chloral = 147.5 g/mol
Now,
According to the stoichiometry,
We require 2 moles of chlorobenzene and a mole of chloral to form a mole of DDT. Thus, chlorobenzene is the excess reactant.
So, for given masses;
No. of moles of chlorobenzene $=\dfrac{1142g}{112.5g/mol}=10.15mol$
No. of moles of chloral $=\dfrac{485g}{147.5g/mol}=3.29mol$
No. of moles of excess reactant i.s. left reactant after the reaction $=10.15mol-\left( 2\times 3.29mol \right)=3.57mol$
Thus, mass of excess reactant $=3.57mol\times 112.5g/mol=401.63g$
Note:Do note to use formulas and the units properly as some confusion can arise while solving the problem.
Before solving such problems, do note that you must be aware of the concepts of limiting and excess reactants.
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