You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 ml

No worries! We‘ve got your back. Try BYJU‘S free classes today!

200 ml

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

400 ml

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

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:

Join BYJU'S Learning Program

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.