KMnO 4 reacts with oxalic acid according to the equation-2 MnO 4-5 C 2 O 42-16 H - 2 Mn 2-10 CO 2-8 H 2 OHere- 20 mL of 0-1 M KM nO 4 is equivalent to-A- 120 mL of 0-25 M H 2 C 2 O 4B- 150 mL of 0-1 M H 2 C 2 O 4C- 50 mL of 0-10 M H 2 C 2 O 4D- 50 mL of 0-20 M H 2 C 2 O 4

The correct option is C 50 mL of 0.10 M nbsp;H_2C_2O_4The ratio of N_KMnO_4 to N_H_2C_2O_4 is 2/5 as given in the equation 2/5=(M× V)_KMnO_4/(M× V)_H_2C_ ...

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Byju's Answer

Standard XII

Chemistry

Equivalent Mass

KMnO 4 reacts...

Question

$K M n O_{4}$ reacts with oxalic acid according to the equation:
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} ⟶ 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$
Here, 20 mL of 0.1 M $K M n O_{4}$ is equivalent to:

120 mL of 0.25 M

H_{2} C_{2} O_{4}

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150 mL of 0.1 M

H_{2} C_{2} O_{4}

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50 mL of 0.10 M

H_{2} C_{2} O_{4}

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50 mL of 0.20 M

H_{2} C_{2} O_{4}

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Solution

The correct option is C 50 mL of 0.10 M $H_{2} C_{2} O_{4}$
$T h e r a t i o o f N_{K M n O_{4}} t o N_{H_{2} C_{2} O_{4}} i s \frac{2}{5} a s g i v e n i n t h e e q u a t i o n$
$\frac{2}{5} = \frac{(M \times V)_{K M n O_{4}}}{(M \times V)_{H_{2} C_{2} O_{4}}}$
$(M \times V_{(m L)})_{H_{2} C_{2} O_{4}} = 5$
so, for option C
$50 \times 0.10 = 5$

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Similar questions

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:

M n O_{4}^{-}

reacts with oxalate 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 m L

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

is equivalent to -

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 m L

of 0.1 M

K M n O_{4}

is equivalent to:

M n O_{4}^{-}

reacts with oxalate 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 m L

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

is equivalent to -

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

20 mL of 0.1 M

K M n O_{4}

will react with?

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KMnO4 reacts with oxalic acid according to the equation: 2MnO−4+5C2O2−4+16H+⟶2Mn2++10CO2+8H2O Here, 20 mL of 0.1 M KMnO4 is equivalent to:

The correct option is C 50 mL of 0.10 M H2C2O4The ratio of NKMnO4 to NH2C2O4 is 25 as given in the equation 25=(M×V)KMnO4(M×V)H2C2O4 (M×V(mL))H2C2O4=5 so, for option C 50×0.10=5

$K M n O_{4}$ reacts with oxalic acid according to the equation:
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} ⟶ 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$
Here, 20 mL of 0.1 M $K M n O_{4}$ is equivalent to: