
Number of moles of $Mn{O_4}^ - $required to oxidize one mole of ferrous oxalate
completely in the acidic medium will be:
A. $7.5\;moles$
B. $0.2\;moles$
C. $0.6\;moles$
D. $0.4\;moles$
Answer
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Hint: As we know that, the mole is the unit of measurement of the physical quantity designated as amount of substance. In practice, one mole of any kind of chemical species is about \[6.022 \times {10^{23}}\] atoms.
Step by step answer:Ferrous ion will always get oxidized to ferrate, \[ + 2, + 3\]that means it losses one \[{e^\_}\], oxalate ion gets to carbon dioxide, carbon changes its oxidation number from \[ + 3\;, + 4\],but oxalate ion contains \[2{e^\_}\], so it loses \[2{e^\_}\] that means \[1\] moles ferrous oxalate is going to lose \[3{e^\_}\].
If we consider \[Mn{O_4}^ - \] is in its \[7\] ionic state, and since the total charge of \[Mn{O_4}\] is\[ - 1\].
Number of moles of \[Mn{O_4}^ - \] required to oxidize one mole of \[Fe{C_2}{O_4}\] completely in an acidic medium. We see that\[5\;mol\]. Of \[Fe{C_2}{O_4}\] gives \[3\] mole of\[Mn{O_4}^ - \]. But when \[1\] mole of \[Fe{C_2}{O_4}\] is oxidized then it gives \[\dfrac{3}{5}\] moles of\[Mn{O_4}^ - \], it means \[0.6\] moles of\[Mn{O_4}^ - \].
\[3Mn{O^ - }_4 + 24{H^ + } + 5Fe{C_2}{O_4} \to 3M{n^{ + 2}} + 12{H_2}O + 10C{O_2} + 5F{e^{ + 3}}\]
\[
5\;moles\;of\;Fe{C_2}{O_4} \to 3\;moles\;of\;Mn{O^ - }_4\\
\Rightarrow 1\;moles\;of\;Fe{C_2}{O_4} \to \dfrac{3}{5}\;moles\;Mn{O^ - }_4\\
\Rightarrow 1\;moles\;of\;Fe{C_2}{O_4} = \;0.6\;moles\;of\;Mn{O_4}^ -
\]
The mole concept is not necessarily restricted to material entities. We can speak of a mole of chemical reactions and so on. The number \[6.022 \times {10^{23}}\] is known as Avogadro number, it should be designated as Avogadro’s constant and expressed as \[mo{l^{ - 1}}\]
\[1\] Moles of atoms or ions or molecules = atomic or molecular mass of it in grams or molar mass.
Therefore, the correct answer is C.
Note: Mole concept is very important and very useful. It includes boyle’s law, Charles law, Gay-Lussac law, principle of atomic conversion, limiting reagents, molarity, molality, mole fraction, normality, etc.
Step by step answer:Ferrous ion will always get oxidized to ferrate, \[ + 2, + 3\]that means it losses one \[{e^\_}\], oxalate ion gets to carbon dioxide, carbon changes its oxidation number from \[ + 3\;, + 4\],but oxalate ion contains \[2{e^\_}\], so it loses \[2{e^\_}\] that means \[1\] moles ferrous oxalate is going to lose \[3{e^\_}\].
If we consider \[Mn{O_4}^ - \] is in its \[7\] ionic state, and since the total charge of \[Mn{O_4}\] is\[ - 1\].
Number of moles of \[Mn{O_4}^ - \] required to oxidize one mole of \[Fe{C_2}{O_4}\] completely in an acidic medium. We see that\[5\;mol\]. Of \[Fe{C_2}{O_4}\] gives \[3\] mole of\[Mn{O_4}^ - \]. But when \[1\] mole of \[Fe{C_2}{O_4}\] is oxidized then it gives \[\dfrac{3}{5}\] moles of\[Mn{O_4}^ - \], it means \[0.6\] moles of\[Mn{O_4}^ - \].
\[3Mn{O^ - }_4 + 24{H^ + } + 5Fe{C_2}{O_4} \to 3M{n^{ + 2}} + 12{H_2}O + 10C{O_2} + 5F{e^{ + 3}}\]
\[
5\;moles\;of\;Fe{C_2}{O_4} \to 3\;moles\;of\;Mn{O^ - }_4\\
\Rightarrow 1\;moles\;of\;Fe{C_2}{O_4} \to \dfrac{3}{5}\;moles\;Mn{O^ - }_4\\
\Rightarrow 1\;moles\;of\;Fe{C_2}{O_4} = \;0.6\;moles\;of\;Mn{O_4}^ -
\]
The mole concept is not necessarily restricted to material entities. We can speak of a mole of chemical reactions and so on. The number \[6.022 \times {10^{23}}\] is known as Avogadro number, it should be designated as Avogadro’s constant and expressed as \[mo{l^{ - 1}}\]
\[1\] Moles of atoms or ions or molecules = atomic or molecular mass of it in grams or molar mass.
Therefore, the correct answer is C.
Note: Mole concept is very important and very useful. It includes boyle’s law, Charles law, Gay-Lussac law, principle of atomic conversion, limiting reagents, molarity, molality, mole fraction, normality, etc.
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