In the Lane's process for the manufacture of hydrogen the reagents used in oxidation-reduction process are respectively:

F e / H_{2} O

and water gas

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F e_{3} O_{4} a n d H_{2}

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O_{2} a n d H_{2}

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F e / H_{2} O

and

F e_{3} O_{4}

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Solution

The correct option is B $F e / H_{2} O$ and water gas
The Lane's process involves following reactions:-
$3 F e + 4 H_{2} O ⟶ F e_{3} O_{4} + 4 H_{2}$
$F e_{3} O_{4} + C O ⟶ 3 F e + C O_{2}$
water
gas
Net reaction :- $C O + H_{2} O ⟶ C O_{2} + H_{2}$
The iron and steam are used to produce $F e_{3} O_{4}$ & $H_{2}$ which then reacts with water gas to liberate $C O_{2}$ and produce fresh iron which again reacts with steam & the process goes on. It is called Lane's process for manufacturing of hydrogen gas.
Thus the answer is option (A) $F e / H_{2} O$ and water gas.

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Similar questions

H_{2} O (s) + F_{2} (g) \to H F (g) + H O F (g)

F e + H_{2} O \to F e_{3} O_{4} + H_{2}

(F e_{3} O_{4})

8 g

F e = 56

F e + H_{2} O \to F e_{3} O_{4} + 4 H_{2}

H_{2} O

F e

\to

F e_{3} O_{4}

H_{2}

H_{2} O

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