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Standard XII

Chemistry

Oxides of Sulphur

10-2 moles of...

Question

$10^{- 2}$ moles of $F e_{3} O_{4}$ is treated with excess of KI solution in presence of dilute $H_{2} S O_{4}$ , the products are
$F e^{2 +}$ and $I_{2}$ (g). What volume of 0.1 (M) $N a_{2} S_{2} O_{3}$ will be needed to reduce the liberated $I_{2}$ (g) ?

50 ml

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100 ml

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200 ml

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400 ml

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Solution

The correct option is C 200 ml

$∴ F e_{3} O_{4} + 2 I^{-} \to F e^{2 +} + I_{2}$
$∴$ no.of moles of $I_{2}$ produced $= 10^{- 2} m o l e s$
Let v ml $0.1 (M) N a_{2} S_{2} O_{3}$ solution is required
$∴ v \times 10^{- 4} = 2 \times 10^{- 2}$
or v=200 ml.

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Similar questions

Q. A

3.0 g

sample containing

F e_{3} O_{4}, F e_{2} O_{3}

and an inert impure substance is treated with excess of

K I

solution in presence of dilute

H_{2} S O_{4}

. The entire iron is converted to

F e^{2 +}

along with the liberation of iodine. The resulting solution is diluted to

100 m L

. A

20 m L

of the dilute solution requires

11.0 m L

0.5 M N a_{2} S_{2} O_{3}

solution to reduce the iodine present. A

50 m L

of the diluted solution after complete extraction of iodine requires

12.8 m L

0.25 M K M n O_{4}

solution in dilute

H_{2} S O_{4}

medium for oxidation of

F e^{2 +}

. Calculate the percentage of

F e_{2} O_{3}

and

F e_{3} O_{4}

in the original sample.

Q. A mixture of 0.02 mole of

K B r O_{3}

and 0.01 mole KBr was treated with excess of KI and acidified. Determine the volume of 0.1 M

N a_{2} S_{2} O_{3}

solution required to consume the liberated iodine.

Q. A solution of

0.2 g

of a compound containing

C u^{2 +}

and

C_{2} O_{4}^{2 -}

ions on titration with

0.02 M

K M n O_{4}

in the presence of

H_{2} S O_{4}

consumes

22.6

mL of the oxidant.

The resultant solution is neutralised with

N a_{2} C O_{3}

, acidified with dilute acetic acid and treated with excess

K I

. The liberated iodine requires

11.3

mL of

0.05

N a_{2} S_{2} O_{3}

solution for complete reduction. The mole ratio of

C u^{2 +}

and

C_{2} O_{4}^{2 -}

in the compound will be :

Q. A 5g sample containing

F e_{3} O_{4} (F e O + F e_{2} O_{3})

and an inert impurity is treated with excess of KI solution in the presence of dilute

H_{2} S O_{4}

. The entire iron converted to Ferrous ion along with the liberation of Iodine. The resulting solution is diluted to 100 mL. 20 mL of the diluted solution requires

10 m L o f 0.5 M N a_{2} S_{2} O_{3}

solution to reduce the iodine present. Calculate approximate percentage of

F e_{2} O_{3}

50

ml of an aqueous solution of

H_{2} O_{2}

were treated with an excess of

K I

solution and dilute

H_{2} S O_{4}

. The liberated iodine required

20

mL of

0.1

N N a_{2} S_{2} O_{3}

solution for complete interaction. The concentration of

H_{2} O_{2}

in g/L is:

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10−2 moles of Fe3O4 is treated with excess of KI solution in presence of dilute H2SO4, the products are Fe2+ and I2 (g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2 (g) ?

The correct option is C 200 ml ∴Fe3O4+2I−→Fe2++I2 ∴ no.of moles of I2 produced =10−2moles Let v ml 0.1(M)Na2S2O3 solution is required ∴v×10−4=2×10−2 or v=200 ml.

$10^{- 2}$ moles of $F e_{3} O_{4}$ is treated with excess of KI solution in presence of dilute $H_{2} S O_{4}$ , the products are
$F e^{2 +}$ and $I_{2}$ (g). What volume of 0.1 (M) $N a_{2} S_{2} O_{3}$ will be needed to reduce the liberated $I_{2}$ (g) ?

The correct option is C 200 ml

$∴ F e_{3} O_{4} + 2 I^{-} \to F e^{2 +} + I_{2}$
$∴$ no.of moles of $I_{2}$ produced $= 10^{- 2} m o l e s$
Let v ml $0.1 (M) N a_{2} S_{2} O_{3}$ solution is required
$∴ v \times 10^{- 4} = 2 \times 10^{- 2}$
or v=200 ml.