For the reaction, Fe 0.95 O (molar mass : M ) − → Fe 2 O 3 . What is the eq. wt. of Fe 0.95 O ?

Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsEquivalent Weight and n-factorMedium2 minQB

ChemistryMediumsingle choice

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$Eq u i v a l e n tw e i g h t = \frac{\frac{M o l ec u l a r w e i g h t}{A t o mi c w e i g h t}}{n - f a c t or}$

The n-factor is very important in both redox and non-redox reactions. With the help of the n-factor, we can predict the molar ratio of the reactant species taking part in reactions. In general, the n-factor of an acid/base is the number of moles of H⁺/OH⁻ furnished per mole of acid/base. The n-factor of a reactant is the number of moles of electrons lost or gained per mole of reactant.

For the reaction, $Fe_{0.95} O$ (molar mass : M $) - \to Fe_{2} O_{3}$ . What is the eq. wt. of $Fe_{0.95} O$ ?

Options:

Answer:

Solution:

The given chemical formula is $Fe_{0.95} O - Fe_{2} O_{3}$ .

The calculation is as follows:

$n.f = (3 - \frac{2}{0.95}) \times 0.95 = 0.85$

$Eq.wt = \frac{M}{0.85}$

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Equivalent Weight and n-factor

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