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Standard XII

Chemistry

Potassium Permanganate

A gas when pa...

Question

A gas when passes through $K_{2} C r_{2} O_{7}$ and dil. $H_{2} S O_{4}$ solution turns it green, the gas is:

C O_{2}

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N H_{3}

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S O_{2}

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C l_{2}

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Solution

The correct option is B $S O_{2}$
When $sulphur dioxide$ gas is passed into acidified dichromate solution, sulphur dioxide is oxidized to sulphuric acid.

Potassium dichromate is reduced to chromic sulphate. The orange solution becomes green.
$K_{2} C r_{2} O_{7} + 3 S O_{2} + H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$

Hence, option $C$ is the correct answer. The gas is $sulphur dioxide$ .

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Similar questions

Q. On treating a compound with warm dil.

H_{2} S O_{4},

gas

X

is evolved which turns

K_{2} C r_{2} O_{7}

paper acidified with dil.

H_{2} S O_{4}

to a green compound

Y .

X

and

Y

respectively are:

Q. (A) + dil.

H_{2} S O_{4} △ ⟶ g a s

(B). Gas (B) turns

K_{2} C r_{2} O_{7} / H^{+}

solution green.
Aqueous solution of (A) +

B a C l_{2} ⟶ w h i t e p p t (C)

Filtrate after removing (C) +

B r_{2}

water

⟶ w h i t e p p t (C)

. Here (A), (B) and (C) are
(A)

S O_{3}^{2 -} + S O_{4}^{2 -}

(B)

S O_{2}

and (C)

B a S O_{4}

A

imparts green color to the flame. Its solution does not give a precipitate on passing

H_{2} S

. When it is heated with solid

K_{2} C r_{2} O_{7}

and conc.

H_{2} S O_{4}

, reddish brown gas is evolved. The gas when passed in an aqueous solution of

N a O H

turns it yellow. Find out

A

Q. A gas

X

occupies

22.4

L at S.T.P. The gas turns acidified

K_{2} C r_{2} O_{7}

green and decolourizes acidified

K M n O_{4}

solution. White turbidity appears when

H_{2} S

gas is passed through aqueous solution of gas, then

X

will be :

Q. When

S O_{2}

is passed through a solution of

K_{2} C r_{2} O_{7}

turns green. The change in oxidation state of chromium and sulphur are ___________________ respectively.

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A gas when passes through K2Cr2O7 and dil. H2SO4 solution turns it green, the gas is:

A gas when passes through $K_{2} C r_{2} O_{7}$ and dil. $H_{2} S O_{4}$ solution turns it green, the gas is: