Question

Mixture X= 0.02 mol of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] and 0.02 mol of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}\] was prepared in 2 L of solution.
1 L of mixture X + excess \[{\text{AgN}}{{\text{O}}_{\text{3}}} \to {\text{Y}}\]
1 L of mixture X + excess \[{\text{BaC}}{{\text{l}}_{\text{2}}} \to {\text{Z}}\]
Number of moles of Y and Z are :
A) 0.03, 0.02
B) 0.01,0.02
C) 0.01,0.01
D) 0.02,0.02

Answer 1

Hint: Using moles of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] and volume of solution calculate its concentration. Similarly, calculate the concentration of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}\] using its moles. Write the balanced reactions of mixture X with an excess of \[{\text{AgN}}{{\text{O}}_{\text{3}}}\] and excess of \[{\text{BaC}}{{\text{l}}_{\text{2}}}\]. Using reaction stoichiometry calculates the number of moles of products X and Y.

Complete step by step answer:
We have given that mixture X contains 0.02 mol of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] and 0.02 mol of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}\]
The volume of solution given to us is 2L.
So, now we can calculate the concentration of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] and \[{\text{(Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br)S}}{{\text{O}}_{\text{4}}}\] in the mixture as follows:
\[{\text{Molarity = }}\dfrac{{{\text{Moles of solute }}}}{{{\text{Volume of solution in litres}}}}\]
To calculate the concentration of \[{\text{Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] substitute 0.02 mol for moles of \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] and 2L for the volume of the solution.

\[[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}] = \dfrac{{{\text{0}}{\text{.02 mol}}}}{{{\text{2L}}}} = 0.01{\text{M}}\]
Similarly, we can calculate the concentration of \[{\text{(Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br)S}}{{\text{O}}_{\text{4}}}\] by substituting 0.02 mol and 2L of solution.
\[[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}] = \dfrac{{{\text{0}}{\text{.02 mol}}}}{{{\text{2L}}}} = 0.01{\text{M}}\]
Now, to calculate the moles of product Y we have to write the balanced reaction of mixture X with an excess of\[{\text{AgN}}{{\text{O}}_{\text{3}}}\].
Out of two complexes in the mixture \[{\text{AgN}}{{\text{O}}_{\text{3}}}\] will react with \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] and will give \[{\text{AgBr }}\] precipitate as follows:
\[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}} + {\text{AgN}}{{\text{O}}_{\text{3}}} \to {\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]N}}{{\text{O}}_{\text{3}}} + {\text{AgBr }} \downarrow \]
1L mixture of 0.01M \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] contain
\[{\text{1L}} \times 0.01{\text{M[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}} = {\text{0}}{\text{.01mol [Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\]
From the balanced reaction, we can say that the mole ratio of complex \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{]Br}}\] and product Y (\[{\text{AgBr }}\]) is 1:1
So, moles of Y = moles of \[{\text{AgBr }}\]= 0.01 mol
Now, similarly, we can calculate the moles of product Z as follows:
Out of two complexes in the mixture \[{\text{BaC}}{{\text{l}}_{\text{2}}}\] will react with \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}\] and will give \[{\text{BaS}}{{\text{O}}_{{\text{4 }}}}\] precipitate as follows:
\[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}} + {\text{BaC}}{{\text{l}}_{\text{2}}} \to {\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]C}}{{\text{l}}_{\text{2}}} + {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{ }} \downarrow \]
1L mixture of 0.01M \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}\] contain
\[{\text{1L}} \times 0.01{\text{M[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}} = {\text{0}}{\text{.01mol [Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}\]
From the balanced reaction, we can say that the mole ratio of complex \[{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{5}}}{\text{Br]S}}{{\text{O}}_{\text{4}}}\]and product Z(\[{\text{BaS}}{{\text{O}}_{{\text{4 }}}}\]) is 1:1
So, moles of Z = moles of \[{\text{BaS}}{{\text{O}}_{{\text{4 }}}}\]= 0.01 mol
Thus, the number of moles of Y and Z are 0.01 mol and 0.01 mol respectively.

Hence the correct option is (C).

Note: Writing the balance reaction is very important in these types of problems as the moles of the products depend on the stoichiometry of the reaction. Molar concentration indicates moles of solute present in a liter of solution.