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Kinetic Theory of Gases

Mixture X of ...

Question

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 :

A

0.01, 0.02

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B

0.02, 0.01

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C

0.01, 0.01

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D

0.02, 0.02

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Solution

The correct option is B 0.01, 0.01
1 L of the mixture contains 0.01 mole of $[C o (N H_{3})_{4} S O_{4}] B r$ and 0.01 mole of $[C o (N H_{3})_{5} B r] S O_{4}$ .
0.01 mole of $[C o (N H_{3})_{4} S O_{4}] B r$ reacts with excess of silver nitrate to produce 0.01 mole of Y which is $A g B r$ .
$[C o (N H_{3})_{4} S O_{4}] B r 0.01 m o l e + A g N O_{3} e x c e s s \to [C o (N H_{3})_{4} S O_{4}] N O_{3} + A g B r Y, 0.01 m o l e$
0.01 mole of $[C o (N H_{3})_{5} B r] S O_{4}$ reacts with excess of $B a C l_{2}$ to give 0.01 mole of Z which is $B a S O_{4}$ .
$[C o (N H_{3})_{5} B r] S O_{4} 0.01 m o l e + B a C l_{2} e x c e s s \to [C o (N H_{3})_{5} B r] C l_{2} + B a S O_{4} Z, 0.01 m o l e$

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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 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

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 :

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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