In the following change, 3Fe+4H_2O→Fe_3O_4+4H_2 If the atomic weight of iron is 56, then its equivalent weight will be

Correct answer: (B) 21. 3Fe^0→Fe3+(8/3)+8e^(-) So E=M/(n-factor)=M/(8/3)=(56×3)/8=21 .

Q43 / 92 · NTA Abhyas (NEET) 2020 · Chemistry · Redox Reactions

In the following change,
$3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$
If the atomic weight of iron is 56, then its equivalent weight will be

3 F e^{0} \to F e_{3}^{+ (8 / 3)} + 8 e^{-}

E = \frac{M}{n - f a c t o r} = \frac{M}{8 / 3} = \frac{56 \times 3}{8} = 21

Hence, the correct answer is option: 'B'.