Question

A sample of $FeS{{O}_{4}}$ , and $Fe{{C}_{2}}{{O}_{4}}$ is dissolved in ${{H}_{2}}S{{O}_{4}}$ . the complete oxidation of sample required $\dfrac{8}{3}eq$ of $KMn{{O}_{4}}$ . After oxidation, the reaction was reduced by Zn. On again oxidation by $KMn{{O}_{4}}$ required $\dfrac{5}{3}eq.$ the mole ratio of $FeS{{O}_{4}}$ , and $Fe{{C}_{2}}{{O}_{4}}$ is:
(A) $\dfrac{3}{7}$
(B) $\dfrac{7}{3}$
(C) $\dfrac{5}{7}$
(D) $\dfrac{7}{5}$

Answer 1

Hint: The number of substances formed in stoichiometric problems can be expressed in moles. In a chemical reaction, mass and the number of atoms must be conserved, the number of molecules not conserved. A conversion factor that relates to the amounts in moles of any substances in a chemical reaction is a mole ratio.

Complete step by step answer:
Mole ratio is determined by observing the coefficients in front of formulas in a balanced chemical equation and this is also called a mole-to-mole ratio.
Lets equivalent of $Fe{{C}_{2}}{{O}_{4}}$ = a
Given, Equivalent of $KMn{{O}_{4}}$ = $\dfrac{8}{3}$
Equivalent of $KMn{{O}_{4}}$ = 2(equivalent of $F{{e}^{+2}}$ )+ equivalent of ${{C}_{2}}{{O}_{4}}^{-2}$
Given, Equivalent of $KMn{{O}_{4}}$ during oxidation = $\dfrac{5}{3}$
Since, Equivalent of $KMn{{O}_{4}}$ = equivalent of $F{{e}^{+3}}$
Let, equivalent of $F{{e}^{+3}}$ = b = equivalent of $FeS{{O}_{4}}$
Then, from the above consideration, we can obtain two relations,
$\Rightarrow \dfrac{8}{3}=2a+b$ -- (1)
$\Rightarrow \dfrac{5}{3}=a+b$ -- (2)
By solving the above two equations, a = 1, and b =$\dfrac{2}{3}$
Since, a = equivalent of $Fe{{C}_{2}}{{O}_{4}}$ , and b = equivalent of $FeS{{O}_{4}}$
Hence, the mole ratio of $FeS{{O}_{4}}$ , and $Fe{{C}_{2}}{{O}_{4}}$ is $\dfrac{7}{3}$
So, the correct answer is “Option B”.

Note: In many chemistry problems, the mole ratio is used as a conversion factor between reactants and products. If the mole ratio is unbalanced, there will be a leftover reactant in a chemical reaction, which we cannot calculate the mole ratio for an unbalanced chemical reaction. mole ratio which helps to convert between quantities of compound in a chemical reaction and with the help of mole ratio easily understand about moles of reactants to moles of products using coefficients.