100 mL solution of ferric alum- Fe2SO43- NH42SO4-24H2O molecular weight - 964 g-mol containing 2-41 g of salt was boiled with Fe when the reaction- Fe - Fe2SO43- 3 FeSO4- takes place- The unreacted iron was filtered off and the solutions was titrated with M60 K2Cr2O7 in acidic medium- Moles of FeSO4 formed when Cu reacts with Fe2SO43 is -

The correct option is A 0.0075 Fe+1 #160;molFe_2(SO_4)_3→3 #160;mol3FeSO_4 1 mol of Fe_2(SO_4)_3.(NH_4)_2SO_4.24H_2O= 1 mol Fe_ ...

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Byju's Answer

Standard XII

Chemistry

Preparation and Properties and Uses of Potassium Hydroxide

100 mL soluti...

Question

$100$ mL solution of ferric alum, $F e_{2} (S O_{4})_{3} . (N H_{4})_{2} S O_{4} .24 H_{2} O$ (molecular weight $= 964$ g/mol) containing $2.41$ g of salt was boiled with $F e$ when the reaction, $F e + F e_{2} (S O_{4})_{3} \to 3 F e S O_{4}$ , takes place. The unreacted iron was filtered off and the solutions was titrated with $\frac{M}{60} K_{2} C r_{2} O_{7}$ in acidic medium.

Moles of $F e S O_{4}$ formed when $C u$ reacts with $F e_{2} (S O_{4})_{3}$ is :

0.0075

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0.005

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0.001

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Solution

The correct option is A $0.0075$
$F e + F e_{2} (S O_{4})_{3} 1 m o l \to 3 F e S O_{4} 3 m o l$

$1$ mol of $F e_{2} (S O_{4})_{3} . (N H_{4})_{2} S O_{4} .24 H_{2} O = 1$ mol $F e_{2} (S O_{4})_{3}$

$\frac{2.41}{964}$ mol of ferric alum $= 2.41 964$ mol of $F e_{2} (S O_{4})_{3}$

$= \frac{3 \times 2.41}{964}$ $= 0.0075$ mol of $F e S O_{4}$

Therefore, option A is correct.

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Similar questions

25

mL of a solution of ferric alum,

F e_{2} (S O_{4})_{3} . (N H_{4})_{2} . S O_{4} .24 H_{2} O

containing

1.25

g of the salt was boiled with iron when the reaction occurs as:

F e + F e_{2} (S O_{4})_{3} ⟶ 3 F e S O_{4}

The unreacted iron was filtered off and solution treated with

0.107 N

K M n O_{4}

in acid medium.

What is titre value of

F e

in mL as nearest integer?

Q. A

10

g mixture of

C u_{2} S

and

C u S

was treated with

200

mL of

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

in acid solution producing

S O_{2}

C u^{2 +}

and

M n^{2 +}

. The

S O_{2}

was boiled off and the excess of

M n O_{4}^{-}

was titrated with

175

mL of

1 M F e^{2 +}

solution. Calculate the percentage of

C u S

in the original mixture (to nearest integer).

3

moles of a mixture of

F e S O_{4}

and

F e_{2} (S O_{4})_{3}

required

100

mL of

2

K M n O_{4}

solution in acidic medium. Hence, mole fraction of

F e S O_{4}

in the mixture is:

Q. A

10 g

sample of

C u S

and

C u_{2} S

was treated with

100 m L

1.25 M K_{2} C r_{2} O_{7}

to produce

C r^{3 +}, C u^{2 +}

and

S O_{2}

. The excess oxidant was reacted with

50 m L

F e^{2 +}

solution.

25 m L

of the same

F e^{2 +}

solution required

0.875 M K M n O_{4}

under acidic condition, the volume of

K M n O_{4}

used was

20 m L

. Find the percentage of

C u_{2} S

in the sample.
Given: molar mass of

C u

63.5 g m o l^{- 1}

Q. 1 gm of an iron are containing 50% ferrous

(F e^{2 +})

and ferric ion

(F e^{3 +})

and rest 50% impurities was dissolved in concentrated hydrochloric acid and the filtered solution was raised to 100 ml in flask. 50 ml of the solution were treated with M/10

K_{2} C r_{2} O_{7}

, which give titre value of 5 ml. Find the percentage of ferric ion in the ore

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The correct option is A 0.0075Fe+Fe2(SO4)31mol→3FeSO43mol1 mol of Fe2(SO4)3.(NH4)2SO4.24H2O=1 mol Fe2(SO4)32.41964 mol of ferric alum =2.41964 mol of Fe2(SO4)3=3×2.41964 =0.0075 mol of FeSO4Therefore, option A is correct.