KMnO 4 reacts with ferrous ammonium sulphate according to the equation MnO 4-5 F e 2-8 H - Mn 2-5 F e 3-4 H 2 O- here 10 mlof 0-1 M KM nO 4 is equivalent toA- 20 ml of 0-1 M F eSO 4B- 30 ml of 0-1 M F eSO 4C- 40 ml of 0-1 M F eSO 4D- 50 ml of 0-1 M F eSO 4

The correct option is D 50 ml of 0.1 M FeSO_4KMnO_4 = Mohr salt M_1 V_1/n_1 = M_2 V_2/n_2; 0.1 × 10/1=M_2 V_2/5; M_2 V_2 = 5 The option that has the same numb ...

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Byju's Answer

Standard IX

Chemistry

Equivalent Mass

KMnO 4 reacts...

Question

$K M n O_{4}$ reacts with ferrous ammonium sulphate according to the equation $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ , here 10 ml of 0.1 M $K M n O_{4}$ is equivalent to

20 ml of 0.1 M

F e S O_{4}

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30 ml of 0.1 M

F e S O_{4}

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40 ml of 0.1 M

F e S O_{4}

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50 ml of 0.1 M

F e S O_{4}

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Solution

The correct option is D 50 ml of 0.1 M $F e S O_{4}$
$K M n O_{4} = M o h r s a l t$
$\frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n_{2}}; \frac{0.1 \times 10}{1} = \frac{M_{2} V_{2}}{5}; M_{2} V_{2} = 5$

The option that has the same number of moles will be the answer.

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Similar questions

K M n O_{4}

reacts with ferrous sulphate according to the following equation:

M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

here,

10 m L

0.1 M K M n O_{4}

is equivalent to

[C P M T 1999]

$K M n O_{4}$ reacts with ferrous ammonium sulphate according to the equation $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ , here 10 ml of 0.1 M $K M n O_{4}$ is equivalent to

Q. In the following redox reaction:

5 {F e}^{2 +} + M n O_{4}^{-} + 8 H^{+} ⇌ {M n}^{2 +} + 5 {F e}^{3 +} + 4 H_{2} O

Given:

{F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}, {E_{1}}^{o}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {M n}^{2 +} + 4 H_{2} O, {E_{2}}^{o}

Potential at the equivalence point is :

K M n O_{4}

reacts with oxalic acid according to the equation,

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

Here, 20 mL of 0.1 M

K M n O_{4}

is equivalent to:

Q. 50 mL aqueous solution of

F e S O_{4}

required 12 mL of 0.2 M

K M n O_{4}

in acidic medium for complete oxidation. Calculate the molarity of ferrous sulphate solution. The reaction is:

M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

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KMnO4 reacts with ferrous ammonium sulphate according to the equation MnO−4+5Fe2++8H+→Mn2++5Fe3++4H2O, here 10 ml of 0.1 M KMnO4 is equivalent to

The correct option is D 50 ml of 0.1 M FeSO4KMnO4=Mohr salt M1V1n1=M2V2n2;0.1×101=M2V25;M2V2=5 The option that has the same number of moles will be the answer.

$K M n O_{4}$ reacts with ferrous ammonium sulphate according to the equation $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ , here 10 ml of 0.1 M $K M n O_{4}$ is equivalent to

The correct option is D 50 ml of 0.1 M $F e S O_{4}$
$K M n O_{4} = M o h r s a l t$
$\frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n_{2}}; \frac{0.1 \times 10}{1} = \frac{M_{2} V_{2}}{5}; M_{2} V_{2} = 5$

The option that has the same number of moles will be the answer.