Question

$100{\text{ }}mL$ solution of ferric alum, $\;F{e_2}{\left( {S{O_4}} \right)_3}.\left( {N{H_4}} \right)S{O_4}.24{H_2}O$ (molecular weight= $964{\text{ }}g/mol$) containing $2.41{\text{ }}g$ of salt was boiled with $Fe$ when the reaction, $Fe{\text{ }} + {\text{ }}F{e_2}{\left( {S{O_4}} \right)_3}\, \to \;\;3FeS{O_4}$ , takes place. The unreacted iron was filtered off and the solutions was titrated with $\dfrac{M}{{60}}{\text{ }}{K_2}C{r_2}{O_7}$ in acidic medium.
Moles of $FeS{O_4}$ formed when $Cu$ reacts with $F{e_2}{\left( {S{O_4}} \right)_3}$ is:
$A.\,0.0075$
$B.\,0.005$
$C.\,0.001$
$D.\,0.002$

Answer 1

Ferric ammonium sulfate or ammonium iron (III)sulfate or ferric alum is a double salt in the class of alums. The ferric alum consists of compounds with the general formula $AB{(S{O_4})_2}\,.\,\,12{\text{ }}{H_2}O$ . It has the appearance of faint purple octahedral crystals. There have been discussions regarding the origin of the colour of crystals, with some attributing it to impurities in the compound, and others claiming that it is a property of the crystal itself. FAS is paramagnetic, acidic and toxic to microorganisms. It is a weak oxidant, capable of being reduced to Mohr's salt, ferrous ammonium sulfate.

Complete step-by-step answer:
$Fe{\text{ }} + {\text{ }}F{e_2}{\left( {S{O_4}} \right)_3}\, \to \;\;3FeS{O_4}$
From this equation we can able to notice that one mole of ferric sulfate ( $F{e_2}{\left( {S{O_4}} \right)_3}\,$ ) gives the three moles of ferrous sulfate ( $FeS{O_4}$ ).
Therefore,
$\;1\,mole\,\,of\,F{e_2}{\left( {S{O_4}} \right)_3}\,\, = \,3\,moles\,of\,FeS{O_4}$
So, the one mole of ferric ammonium sulfate (ferric alum) will be equal to one mole of ferric sulfate; Therefore,
$1\,mole\,of\,\;F{e_2}{\left( {S{O_4}} \right)_3}.\left( {N{H_4}} \right)S{O_4}.24{H_2}O\,\, = \,\,1\,mole\,\,of\,F{e_2}{\left( {S{O_4}} \right)_3}\,$
To calculate the number of moles;
The given mass of the compound is divided by its molecular mass, so that the number of moles can be obtained.
Let’s use the equation;
Number of moles, $n\, = \,\dfrac{{mass}}{{molecular\,mass}}$
The given mass of ferric ammonium sulfate (ferric alum) ( $\;F{e_2}{\left( {S{O_4}} \right)_3}.\left( {N{H_4}} \right)S{O_4}.24{H_2}O$ )=$2.41{\text{ }}g$
The molecular mass of ferric ammonium sulfate (ferric alum) ( $\;F{e_2}{\left( {S{O_4}} \right)_3}.\left( {N{H_4}} \right)S{O_4}.24{H_2}O$ ) = $964{\text{ }}g/mol$
Substitute the values in above equation,
$n\, = \,\dfrac{{2.41{\text{ }}g}}{{964\,g/mol}}$
So, the number of moles will be;
$n\, = \,0.0025$
Or else;
We can write as given in below;
$\dfrac{{2.41{\text{ }}g}}{{964\,g/mol}}$ mol of ferric alum = $\dfrac{{2.41{\text{ }}g}}{{964\,g/mol}}$ mol of $F{e_2}{\left( {S{O_4}} \right)_3}\,$
So, to calculate the number of moles for $FeS{O_4}$ , we need to multiply by $3$;
Therefore,
$= \,\dfrac{{3\, \times \,2.41{\text{ }}g}}{{964\,g/mol}}$
So, the answer we obtain is;
$= \,0.0075$ mol of $FeS{O_4}$

So, the correct answer is $A.\,0.0075$ .

Note: A mole is a unit measurement for the amount of substance in the international system of units i.e., SI unit. A mole of a particle or a mole of a substance is defined as $6.02214076 \times {10^{23}}$ of a chemical unit, that can be ions, atoms, molecules, etc. Originally it was defined as the number of atoms in $12{\text{ }}g$ of carbon-12.