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# The molar ratio of $F{e^{2 + }}$to $F{e^{3 + }}$ in a mixture of $FeS{O_4}$ and $F{e_2}{(S{O_4})_3}$ having equal number of sulphate ion in both ferrous and ferric sulphate is :A. 1:2B. 3:2C. 3:1D. can't be determined

Hint: From the chemical formula of $FeS{O_4}$ and $F{e_2}{(S{O_4})_3}$ it is clear that one molecule $FeS{O_4}$ contains 1$F{e^{2 + }}$and 1$S{O_4}^{2 - }$ ion. And one molecule $F{e_2}{(S{O_4})_3}$ contains 2$F{e^{3 + }}$and 3$S{O_4}^{2 - }$ ions. Molar ratio can be derived from the coefficient from the chemical reaction equation.
According to the chemical formula it is clear that 1 mole $FeS{O_4}$ and $F{e_2}{(S{O_4})_3}$ contains sulfate ions 1 moles and 3 moles respectively.
Now if they have equal number of sulfate ion then amount of $FeS{O_4}$in the mixture should be 3 mole and amount of $F{e_2}{(S{O_4})_3}$ in the mixture should be 1 mole.
$\dfrac{{F{e^{2 + }}}}{{F{e^{3 + }}}} = \dfrac{{3 \times 1}}{{1 \times 2}} \\ or,\dfrac{{F{e^{2 + }}}}{{F{e^{3 + }}}} = \dfrac{3}{2} \\$
The molar ratio of $F{e^{2 + }}$ to $F{e^{3 + }}$ in a mixture of $FeS{O_4}$ and $F{e_2}{(S{O_4})_3}$ having equal number of sulphate ion in both ferrous and ferric sulphate is ${\text{3:2}}$.
Remember that the metal charge in $FeS{O_4}$ is $F{e^{2 + }}$ which is called ferrous ion . And the metal charge in $F{e_2}{(S{O_4})_3}$ is $F{e^{3 + }}$ which is called ferric ion.