Calculate the gram equivalent weight for the indicated reaction.
$F e^{+ 2} ⟶ {F e}^{3 +}$

56 g

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112 g

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28 g

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None of these

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Solution

The correct option is A $56 g$
${F e}^{2 +} ⟶ {F e}^{3 +}, n f a c t o r = 1.$

Molecular weight is equal to the equivalent weight.

$H e n c e, E q . w t . = 56 g$

So, the correct option is (A)

Suggest Corrections

Similar questions

H_{2} S ⟶ S

{F e}_{3} O_{4}

{F e}_{3} O_{4} + {K M n O}_{4} \to {F e}_{2} O_{3} + {M n O}_{2}

H_{2} O_{2} (⟶ O_{2})

{F e}_{2} S_{3}

{F e}_{2} S_{3}

{F e}_{3} S_{3} + O_{2} ⟶ {F e S O}_{4} + {S O}_{2}

F e^{3 +} + 3 e^{-} ⇌ F e, E^{\circ}

- 0.036 v o l t

F e^{3 +} + e^{-} ⇌ F e^{2 +}

0.771 v o l t

E^{\circ}

F e^{2 +} + 2 e^{-} ⇌ F e

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