MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Hydrides of Nitrogen

Mixture X=0...

Question

Mixture $X = 0.02$ 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 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 :

A

0.03, 0.02

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

0.01, 0.02

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

0.01, 0.01

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

0.02, 0.02

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.01, 0.01$
1.0 L of mixture X contain 0.01 mole of each $[(C o (N H_{3})_{5} S O_{4}] B r$ an $[C o (N H_{3})_{5} B r] S O_{4}$ .

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

$[C o (N H 3)_{5} S O_{4}] B r + A g N O_{3} \to [C o (N H_{3})_{5} S O_{4}] N O_{3} + A g B r$ (0.01 mol)

0.01 mol excess

Similarly, in the second reaction, $B a S O_{4}$ (Z) will be precipitated.

Hence option C is correct.

Suggest Corrections

thumbs-up

0

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

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

[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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Hydrides of Nitrogen - Ammonia and Hydrazine

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Hydrides of Nitrogen

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon