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Frankland Halide Reaction

When an equim...

Question

When an equimolar mixture of $C u_{2} S$ and $C u S$ is titrated with $B a (M n O_{4})_{2}$ in acidic medium, the final product contains $C u^{2 +}, S O_{2}$ and $M n^{2 +}$ . If the molecular weight of $C u_{2} S, C u S$ and $B a (M n O_{4})_{2}$ are $M_{1}, M_{2}$ and $M_{3}$ respectively then,

A

equivalent weight of

C u_{2} S = \frac{M_{1}}{8}

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B

equivalent weight of

C u S = \frac{M_{2}}{6}

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C

equivalent weight of

B a (M n O_{4})_{2} = \frac{M_{3}}{5}

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D

C u_{2} S

and

C u S

both have same equivalents in mixture

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Solution

The correct options are
A equivalent weight of $C u_{2} S = \frac{M_{1}}{8}$
B equivalent weight of $C u S = \frac{M_{2}}{6}$
In $C u_{2} S$ :
Change in oxidation state of $S$ $= + 4 - (- 2) = 6$ .
Change in oxidation state of $C u^{+} = (+ 2 \times 2) - (2 \times 1) = 2$ .
So, eq. wt of $C u_{2} S = \frac{M_{1}}{6 + 2 = 8}$ .
In $C u S$ :
Change in oxidation state of $S$ $= + 4 - (- 2) = 6$ .
Change in oxidation state of $C u^{+} = (+ 2 \times 2) - (2 \times 2) = 0$ .
So, eq. wt of $C u_{2} S = \frac{M_{1}}{6 + 0 = 6}$ .
Also, eq. wt. of $B a (M n O_{4})_{2} = \frac{M_{3}}{2 \times 5 = 10}$ .

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Similar questions

Q. When an equimolar mixture of

C u_{2} S

C u S

is titrated with

B a (M n O_{4})_{2}

in acidic medium, the final product contains

C u^{2 +}

S O_{2}

M n^{2 +}

. If the molecular weight of

C u_{2} S

C u S

B a (M n O_{4})_{2}

M_{1}

M_{2}

M_{3}

respectively then:

Q. The mixture of

C u S

(molar weight =

M_{1}

C u_{2} S

(molecular weight =

M_{2}

) is oxidised by

K M n O_{4}

(molecular weight =

M_{4}

) in acidic medium, the product obtained are

C u^{2 +}

S O_{2 +}

. The equivalent weights of

C u S, C u_{2} S

K M n O_{4}

respectively, are :

C u_{2} S

is converted into

{C u}^{2 +}

{S O}_{2}

, then equivalent weight of

{C u}_{2} S

[M = molecular weight of

{C u}_{2} S

]

Q. The equivalent weight of

{C u}_{2} S

(m o l . w t . = M)

in the following reaction is:

{C u}_{2} S + 4 {M n O}_{4} ⟶ {C u}^{2} + {S O}_{2} + {M n}^{2 +}

10

C u_{2} S

C u S

was treated with

200

0.75 M M n O_{4}^{-}

in acid solution producing

S O_{2}

C u^{2 +}

M n^{2 +}

S O_{2}

was boiled off and the excess of

M n O_{4}^{-}

was titrated with

175

1 M F e^{2 +}

solution. Calculate the percentage of

C u S

in the original mixture (to nearest integer).

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