Question

0.02 mole of $\text{ }\!\![\!\!\text{ Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Br]C{{l}_{2}}$ and 0.02 mole of $[\text{Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Cl]\text{S}{{\text{O}}_{4}}$ are present in 200 cc of a solution X. The number of moles of the precipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride respectively.
(A) 0.02, 0.02
(B) 0.01, 0.02
(C) 0.02, 0.04
(D) 0.04, 0.02

Answer 1

Hint: The reaction of both molecules will take place with excess silver nitrate and excess barium chloride respectively. A number of moles is equal to the mass of the solute divided by the molar mass. Also, 1 mole is equal to Avagadro's number.

Complete step by step answer:
-Firstly, we have to write the reactions with both reactants to know that how much moles are formed of a precipitate by $\text{ }\!\![\!\!\text{ Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Br]C{{l}_{2}}$ and $[\text{Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Cl]\text{S}{{\text{O}}_{4}}$.
-The reaction of $\text{ }\!\![\!\!\text{ Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Br]C{{l}_{2}}$with excess of silver nitrate will give:
$\text{ }\!\![\!\!\text{ Co}{{\left( N{{H}_{3}} \right)}_{5}}Br]\text{C}{{\text{L}}_{2}}\text{ + 2AgN}{{\text{O}}_{3}}\text{ }\to \text{ }\!\![\!\!\text{ Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Br]{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}\text{ + 2AgCl (Y)}$
-Here, 1 mole of $\text{ }\!\![\!\!\text{ Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Br]C{{l}_{2}}$reacts with silver nitrate to give 2 moles of AgCl which is a precipitate (Y). So, it is clear that 0.02 mole of $\text{ }\!\![\!\!\text{ Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Br]C{{l}_{2}}$will give:
                                                    $\begin{align}
  & \text{2 }\cdot \text{ 0}\text{.02} \\
 & \text{= 0}\text{.04} \\
\end{align}$
-Similarly, when $[\text{Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Cl]\text{S}{{\text{O}}_{4}}$reacts with an excess of barium chloride then it will give:
$[\text{Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}\text{Cl }\!\!]\!\!\text{ S}{{\text{O}}_{4}}\text{ + BaC}{{\text{l}}_{2}}\to \text{ }\!\![\!\!\text{ Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}\text{Cl }\!\!]\!\!\text{ C}{{\text{l}}_{2}}\text{ + BaS}{{\text{O}}_{4}}\text{ (Z)}$
 -Here, Barium sulphate is a precipitate (Z). In this reaction, 1 mole of $[\text{Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Cl]\text{S}{{\text{O}}_{4}}$ reacts with an excess of barium chloride to give 1 mole of barium sulphate.
-So, 0.02 moles of $[\text{Co}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{5}}Cl]\text{S}{{\text{O}}_{4}}$will give \[1\text{ }\cdot \text{ 0}\text{.02}\] that is equal to 0.02 moles of barium sulphate.

Therefore, option (D) is correct that is 0.04 mole, 0.02 mole.

Note: In the first reaction, the nitrate ion will replace chloride ion and not silver whereas in a second reaction, sulphate ion is replaced by chloride ion and not by the barium ion because of the strong ability to replace.