Question

Mixture $X = 0.02{\text{mol}}$ of $[Co{(N{H_3})_5}S{O_4}]Br$ and $0.02{\text{mol}}$ of $[Co{(N{H_3})_5}Br]S{O_4}$ was prepared in $2{\text{litres}}$ of solution.
$1{\text{litre}}$ of mixture $X$ $ + $ excess $AgN{O_3}$ $ \to Y$,
$1{\text{litre}}$ of mixture $X$ $ + $ excess $BaC{l_2}$ $ \to Z$
Number of moles of $Y$ and $Z$ are:
A.$0.01,0.01$
B.$0.02,0.01$
C.$0.01,0.02$
D.$0.02,0.02$

Answer 1

Hint:$[Co{(N{H_3})_5}S{O_4}]Br$ and $[Co{(N{H_3})_5}Br]S{O_4}$ are known as ionisation isomers. When one reactant in a reaction is in excess then the amount product formed will be according to the reactant which is not in excess.

Complete step by step answer:
First of all let us talk about excess reagent and limiting reagent.
Excess reagent: The reagents which are present in excess amount during the reaction, are known as excess reagent.
Limiting reagents: The reagents which are present in small amounts during the reaction, are known as limiting reagent.
Now let us talk about these reactions which are given in the question.
First reaction which is given in the question is that $1{\text{litre}}$ of mixture $X$ $ + $ excess $AgN{O_3}$ $ \to Y$, where $X = 0.02{\text{mol}}$ of $[Co{(N{H_3})_5}S{O_4}]Br$ and $0.02{\text{mol}}$ of $[Co{(N{H_3})_5}Br]S{O_4}$ in $2{\text{litres}}$ of solution. Now when we want $1{\text{litre}}$ of mixture $X$ then $X = 0.01{\text{mol}}$ of $[Co{(N{H_3})_5}S{O_4}]Br$ and $0.01{\text{mol}}$ of $[Co{(N{H_3})_5}Br]S{O_4}$ (i.e. mass of the substance in the reaction should remain constant). Now we know that $[Co{(N{H_3})_5}S{O_4}]Br$ will react with $AgN{O_3}$ to form a precipitate of $AgBr$ and $[Co{(N{H_3})_5}Br]S{O_4}$ will remain unreacted. And the reaction as follows:
$[Co{(N{H_3})_5}S{O_4}]Br + AgN{O_3} \to [Co{(N{H_3})_5}S{O_4}]N{O_3} + AgBr$ as amount of $[Co{(N{H_3})_5}S{O_4}]Br$ is $0.01{\text{mol}}$ so the amount of silver bromide will be $0.01{\text{mol}}$ because one mole of $[Co{(N{H_3})_5}S{O_4}]Br$ produce one mole of silver bromide i.e. $AgBr$. So the number of moles of $Y$ is $0.01{\text{mol}}$.
Similarly, second reaction given in the question is that $1{\text{litre}}$ of mixture $X$ $ + $ excess $BaC{l_2}$ $ \to Z$, where $X = 0.02{\text{mol}}$ of $[Co{(N{H_3})_5}S{O_4}]Br$ and $0.02{\text{mol}}$ of $[Co{(N{H_3})_5}Br]S{O_4}$ in $2{\text{litres}}$ of solution. Now when we want $1{\text{litre}}$ of mixture $X$ then $X = 0.01{\text{mol}}$ of $[Co{(N{H_3})_5}S{O_4}]Br$ and $0.01{\text{mol}}$ of $[Co{(N{H_3})_5}Br]S{O_4}$ (i.e. mass of the substance in the reaction should remain constant). Now we know that $[Co{(N{H_3})_5}Br]S{O_4}$ will react with $BaC{l_2}$ to form a precipitate of $BaS{O_4}$ and $[Co{(N{H_3})_5}S{O_4}]Br$ will remain unreacted. And the reaction as follows:
$[Co{(N{H_3})_5}Br]S{O_4} + BaC{l_2} \to [Co{(N{H_3})_5}Br]C{l_2} + BaS{O_4}$ as amount of $[Co{(N{H_3})_5}Br]S{O_4}$ is $0.01{\text{mol}}$ so the amount of barium sulphate will be $0.01{\text{mol}}$ because one mole of $[Co{(N{H_3})_5}Br]S{O_4}$ produce one mole of barium sulphate i.e. $BaS{O_4}$. So the number of moles of $Z$ is $0.01{\text{mol}}$. So the number of moles of $Y$ and $Z$ formed are $0.01,0.01$.

Hence option A is correct.

Note:
Isomerism: It is the phenomenon in which more than one compounds have the same chemical formula but different structures. $[Co{(N{H_3})_5}S{O_4}]Br$ and $[Co{(N{H_3})_5}Br]S{O_4}$ are known as ionisation isomers. Ionisation isomers are those isomers in which molecules are the same i.e. have the same number of atoms present in the atom but the only difference is that they give different ions on ionisation.