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Question: When ferrous oxalate is titrated against \( {K_2}C{r_2}{O_7}{\text{ m}} \) eq. of \( F{e^{2 + }},{C_...

When ferrous oxalate is titrated against K2Cr2O7 m{K_2}C{r_2}{O_7}{\text{ m}} eq. of Fe2+,C2O42F{e^{2 + }},{C_2}O_4^{2 - } and Cr2O72C{r_2}O_7^{2 - } in the redox reaction are X,Y and ZX,Y{\text{ and Z}} respectively, then:
(A) X=YX = Y
(B) X+Y=ZX + Y = Z
(C) X+2Y=ZX + 2Y = Z
(D) 2X+6Y+6Z2X + 6Y + 6Z

Explanation

Solution

A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution. Typically, the titrant (the known solution) is added from a burette to a known quantity of the analyte (the unknown solution) until the reaction is complete.

Complete answer:
Let us firstly write down the chemical reaction for this question and after that we will proceed further. The chemical reaction is:
FeC2O4+K2Cr2O7Fe+2+C2O42+Cr2O72Fe{C_2}{O_4} + {K_2}C{r_2}{O_7} \to F{e^{ + 2}} + {C_2}O_4^{2 - } + C{r_2}O_7^{2 - }
Now if we look at this reaction it is completely balanced except for the potassium ion as we don’t need it in our reaction and in solving the questions. So we will neglect the presence of potassium as if it is where there then the reaction would still be balanced. So if the ferrous oxalate is titrated against K2Cr2O7 m{K_2}C{r_2}{O_7}{\text{ m}} off Fe2+,C2O42F{e^{2 + }},{C_2}O_4^{2 - } and Cr2O72C{r_2}O_7^{2 - } in the redox reaction then let them be X,Y and ZX,Y{\text{ and Z}} respectively. Now here we can see that X=YX = Y from the reaction as their coefficients are the same.
So our answer will be Option A.

Note:
The stoichiometric coefficient is the number written in front of atoms, ions, and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred.