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Percentage Composition

Mixture X c...

Question

Mixture $X$ containing $0.02$ mol of $[C o (N H_{3})_{5} S O_{4}] B r$ and $0.02$ mol of $[C o (N H_{3})_{5} B r] S O_{4}$ was prepared in $2$ L solution.
$1$ L of mixture $X$ + excess of $A g N O_{3}$ solution $\to Y$
$1$ L of mixture $X$ + excess of $B a C l_{2}$ solution $\to Z$
Number of moles of $Y$ and $Z$ is, respectively :

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 $0.01, 0.01$
$1$ L of mixture $X$ contains $\frac{1}{2} \times 0.02 = 0.01$ moles of $[C o (N H_{3})_{5} S O_{4}] B r$ .
It will react with excess of silver nitrate to produce $0.01$ mole of $A g B r$ which is compound $Y$ .
The reaction is as follows:
$[C o (N H_{3})_{5} S O_{4}] B r + A g N O_{3} \to [C o (N H_{3})_{5} S O_{4}] N O_{3} + A g B r$
$1$ L of mixture $Y$ contains $\frac{1}{2} \times 0.02 = 0.01$ moles of $[C o (N H_{3})_{5} B r] S O_{4}$ .
It will react with excess of barium chloride to produce $0.01$ mole of barium sulfate which is compound $Z$ .
The reaction is as follows:
$[C o (N H_{3})_{5} B r] S O_{4} + B a C l_{2} \to [C o (N H_{3})_{5} B r] C l_{2} + B a S O_{4}$

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Similar questions

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

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

[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

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 :

X = 0.02 m o l

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

0.02 m o l

[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

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