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Characteristics of Equilibrium Constant

Mixture X=0.0...

Question

Mixture $X = 0.02 m o l$ of $[C o (N H_{3})_{5} S O_{4}] B r$ and $0.02 m o l$ of $[C o (N H_{3})_{5} B r] S O_{4}$ was prepared in 2 litre of solution.
$1 litre of mixture X+ excess A g N O_{3} \to Y$
$1 litre of mixture X + excess B a C l_{2} \to Z$
Number of moles of Y and Z are

A

0.01, 0.01

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B

0.02, 0.01

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C

0.01, 0.02

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D

0.02, 0.02

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Solution

The correct option is A $0.01, 0.01$
1.0 L of mixture X contain 0.01 mole of each $[C o (N H_{3})_{5} S O_{4}] B r$ and $[C o (N H_{3})_{5} B r] S O_{4}$
Also, with $A g N O_{3}$ , only $[C o (N H_{3})_{5} S O_{4}] B r$ reacts to give $A g B r$ precipitate as:
$[C o (N H_{3})_{5} S O_{4}] B r 0.01 m o l + A g N O_{3} excess \to [C o (N H_{3})_{5} S O_{4}] N O_{3} + A g B r 0.01 m o l$
Similarly, in the second reaction $B a S O_{4}$ will be precipitated
$[C o (N H_{3})_{5} B r] S O_{4} 0.01 m o l + B a C l_{2} excess \to [C o (N H_{3})_{5} S O_{4}] C l_{2} + B a S O_{4} 0.01 m o l$
Option (a) is correct.

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Similar questions

X

0.02

[C o (N H 3)_{5} S O_{4}] B r

[C o (N H 3)_{5} B r] S O_{4}

was prepared in 2 litre of solution.
1 litre of mixture

X +

A g N O 3 \to

Y.
1 litre of mixture

X +

B a C 1_{2} \to Z

Number of moles of

Y

Z

[X] = 0.02

[C o (N H_{3})_{5} S O_{4}] B r

and 0.02 mol of

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in 2 litre of solution.
1 litre of mixture [X] + excess

A g N O_{3} ⟶

[Y]
1 litre of mixture [X] + excess

B a C l_{2} ⟶

[Z]
No. of moles of [Y] and [Z] are :

X = 0.02

[C o (N H_{3})_{5} S O_{4}] B r

and 0.02 mol of

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in 2L of solution.

1L of mixture X + excess

A g N O_{3} \to Y

1L of mixture X + excess

B a C l_{2} \to Z

Number of moles of Y and Z are :

X

0.02

[C o (N H_{3})_{5} S O_{4}] B r

0.02

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in

2

1

X

A g N O_{3}

\to Y

1

X

B a C l_{2}

\to Z

Number of moles of

Y

Z

is, respectively :

Q. Mixture X of 0.02 mole of

[C o (N H_{3})_{4} S O_{4}] B r

and 0.02 mole of

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in 2 litre of solution:

1 litre of mixture X + excess of

A g N O_{3} \to Y

1 litre of mixture X + excess of

B a C l_{2} \to Z

Number of moles of Y and Z respectively are :

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