Question

To measure the quantity of $MnC{l_2}$ dissolved in an aqueous solution, it was completely converted into $KMn{O_4}$ using the reaction, $MnC{l_2} + {K_2}{S_2}{O_8} + {H_2}O\xrightarrow{{}}KMn{O_4} + {H_2}S{O_4} + HCl$ (equation not balanced). Few drops of concentrated $HCl$ were added to this solution and gently warmed. Further, oxalic acid $\left( {225\,g} \right)$ was added in portions till the colour of the permanganate ion disappeared. The quantity of $MnC{l_2}$ (in mg) present in the initial solution:
(Atomic weight in $g\,mo{l^{ - 1}}\,:Mn = 55,Cl = 35.5$)

Answer 1

We can keep the unknown quantity as x. Using the equivalent weight, known weight of oxalic acid and number of electrons transferred in the overall reaction. The mass of $MnC{l_2}$ can be calculated. We know that in chemical reaction electrons are either lost (or) gained. Oxidation number gives the exact number of electrons, which are either lost or gained to form a chemical bond with other species.

Complete step by step answer:
Manganese chloride reacts with potassium persulfate in aqueous solution to form potassium permanganate, sulfuric acid and hydrogen chloride. Oxalic acid is added to potassium permanganate to give manganese ion and carbon dioxide.
For the given reaction, we can write the chemical equation as,

$$MnC{l_2} + {K_2}{S_2}{O_8} + {H_2}O\xrightarrow{{}}KMn{O_4} + {H_2}S{O_4} + HCl \\\ KMn{O_4} + {C_2}{H_2}{O_4}\xrightarrow{{}}M{n^{2 + }} + C{O_2} \\\$$

The above reaction is a redox reaction that is oxidation and reduction takes simultaneously.
We can say the number of $meq$ of $MnC{l_2}$ is equal to the number of $meq$ of $KMn{O_4},$ which is again equal to number of $meq$ of ${\left( {COOH} \right)_2}.$
We know that the oxidation state of $Mn$ in $KMn{O_4}$ is $+ 7$ and the oxidation state of $Mn$ in $MnC{l_2}$ is ${\text{ + 2}}{\text{.}}$
We know the oxidation state of $C$ in $C{O_2}$ is $+ 4.$
The number of electrons changed per mole from $Mn{O_4}^ -$ to $M{n^{2 + }}$ is ${\text{5}}{\text{.}}$ The number of electrons changed per mole from ${\left( {COOH} \right)_2}$ to $C{O_2}$ is ${\text{ + 2}}{\text{.}}$
Let x be the quantity of $MnC{l_2}.$
We can the given mass of oxalic acid is $225\,mg.$
The molar mass of oxalic acid $90\,g\,mo{l^{ - 1}}.$
The molar mass of $MnC{l_2}$ is $126\,g\,mo{l^{ - 1}}.$
We can calculate the quantity of $MnC{l_2}$ as,
Equivalents of $MnC{l_2} =$Equivalents of ${\left( {COOH} \right)_2}$
${\left( {\dfrac{{Unknown\,mass}}{{Molar\,mass}} \times n} \right)_{MnC{l_2}}} = {\left( {\dfrac{{Known\,mass}}{{Molar\,mass}} \times n} \right)_{{C_2}{H_2}{O_4}}}$
Here, we should understand that the value of n represents the number of electrons.
${\left( {\dfrac{x}{{126\,g\,mo{l^{ - 1}}}} \times 5} \right)_{MnC{l_2}}} = {\left( {\dfrac{{225\,mg}}{{90\,g\,mo{l^{ - 1}}}} \times 2} \right)_{{C_2}{H_2}{O_4}}} \\\ x = 126\,mg \\\$
From the above calculation, we can see that the weight of manganese di-chloride $\left( {MnC{l_2}} \right)$ is $126mg.$
Therefore, the quantity of $MnC{l_2}$ present in the initial solution is $126mg.$

Note:
We can calculate the number of electrons changed per mole by the difference in oxidation number of the reactants and the products. We can define the equivalent as the amount of a substance that is reacted with electrons (in one mole) in a redox reaction. Equivalent weight is the mass of one equivalent.