Answer:
0.20 mol
Explanation:
Let's consider the reduction of iron from an aqueous solution of iron (II).
Fe²⁺ + 2 e⁻ ⇒ Fe
The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:
5.6 g × 1 mol/55.85 g = 0.10 mol
2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are
0.10 mol Fe × 2 mol e⁻/1 mol Fe = 0.20 mol e⁻
0.20 mol of electrons is required to deposit 5.6g of iron from a solution of iron (2) tetraoxosulphate(6)
The reduction of iron from an aqueous solution of iron (II).
[tex]Fe^{+2} +2e^{-} \rightarrow Fe[/tex]
The formula for number of moles is as follows:-
[tex]Number \ of \ moles=\frac{Mass}{Molar\ mass}[/tex]
The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:
[tex]5.6 g \times\frac{1\ mol}{55.85\ g} = 0.10 \ mol[/tex]
2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are:-
[tex]0.10 mol Fe\times\frac{2\ mol\ e^{-} }{1\ mol\ e^{-}} = 0.20 \ mol e^{-}[/tex]
Hence, 0.20 mol of electrons is required to deposit 5.6g of iron.
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