The ion An- is oxidised to AO3- by MnO4- changing to Mn2- in acid solution- Given that 2-68 - 10-3 mol of An- requires 1-61 - 10-3 mol of MnO4-What is the weight of one gram equivalent of ACln for the above reaction if the atomic mass of A is 97-

The balanced chemical equation is as given below: ∴ 5A^n+ + (5 n)MnO_4^ → (5 n)Mn^2+ + 5AO_3^ ∴5 mol of Ag^n+/(5 n) mol of MnO_4^ = #160; 2.68 ...

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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

The ion An+...

Question

The ion $A^{n +}$ is oxidised to ${A O_{3}}^{-}$ by ${M n O_{4}}^{-}$ changing to $M n^{2 +}$ in acid solution. Given that $2.68 \times 10^{- 3}$ mol of $A^{n +}$ requires $1.61 \times 10^{- 3}$ mol of ${M n O_{4}}^{-}$ .
What is the weight of one gram equivalent of $A C l_{n}$ for the above reaction if the atomic mass of $A$ is $97$ ?

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Solution

The balanced chemical equation is as given below:
$∴ 5 A^{n +} + (5 - n) {M n O_{4}}^{-} \to (5 - n) M n^{2 +} + 5 {A O_{3}}^{-}$

$∴ \frac{5 mol of A g^{n +}}{(5 - n) mol of {M n O_{4}}^{-}} = \frac{2.68 \times 10^{- 3}}{1.61 \times 10^{- 3}}$

$5 - n = 3$ or $n = 2$
Thus, the charge on $A^{n +}$ is $+ 2$ .
Th oxidation number of $A$ changes from $+ 2$ to $+ 5$ .
The change in oxidation number is $+ 3$ .
The equivalent weight of $A g C l_{2} = \frac{Molar mass}{Change in oxidation number} = \frac{97 + 71}{3} = 56$

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Similar questions

Q. The ion

A^{n +}

is oxidised to

{A O}_{3}^{-}

M n O_{4}^{-}

changing to

{M n}^{2 +}

in acidic solution. Given that

2.68 \times 10^{- 3}

mol pf

A^{n +}

requires

1.61 \times 10^{- 3}

mol of

M n O_{4}^{-}

. The value of

n

is :

K M n O_{4}

oxidises

X^{n + 1}

ion into

{X O}_{3}^{-}

, itself changing to

{M n}^{2 +}

in acid solution.

2.68 \times 10^{- 3}

mol of

X^{n + 1}

requires

1.61 \times 10^{- 3}

mol of

M n O_{4}^{-}

. What is atomic weight of

X

, if the weight of

1

g equivalent of

{X C l}_{n}

56

Q. A solution containing

2.68 \times 10^{- 3} mol

A^{n +}

ions requires

1.61 \times 10^{- 3} mol

M n O_{4}^{-}

for the complete oxidation of

A^{n +}

A O_{3}^{-}

in acidic medium. What is the value of

n

K M n O_{4}

oxidizes

X O^{n +}

ion to

X {_{3}}^{-}

ion, itself changing to

M n^{2 +}

in acid medium.

2.68 \times 10^{- 3}

mole of

X^{n +}

requires

1.61 \times 10^{- 3}

mole of

M n O {_{4}}^{-}

. If the weight of

1

g equivalent of

X C I_{n}

56

, then the atomic mass of X is :

2.68 \times 10^{- 3}

moles of a solution containing an ion

A^{n +}

require

1.61 \times 10^{- 3}

moles

M n O_{4}^{-}

for the oxidation of

A^{n +}

A O_{3}^{-}

in acid medium. What is the value of n?

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The ion An+ is oxidised to AO3− by MnO4− changing to Mn2+ in acid solution. Given that 2.68×10−3 mol of An+ requires 1.61×10−3 mol of MnO4−.What is the weight of one gram equivalent of ACln for the above reaction if the atomic mass of A is 97?