Question

$10\, g$ sample of $FeS$ was roasted and $SO _{2}$ thus produced required $10 \,ml$ of $0.15 \,M\, K _{2} Cr _{2} O _{7}$ solution in acidic medium. Hence the $\%$ of $FeS$ in closest integer in the sample is ............... (Fe $=56, S =32$ )

Answer 1

$4 FeS +7 O _{2}( g ) \longrightarrow 2 Fe _{2} O _{3}+4 SO _{2}( g )$
$Cr _{2} O _{7}{ }^{2-}+2 H ^{+}+3 SO _{2}( g ) \longrightarrow 2 Cr ^{3+}+3 SO _{4}{ }^{2-}+ H _{2} O$
So moles of $K _{2} Cr _{2} O _{7}$ used $=0.15 \times 10^{-2}=1.5 \times 10^{-3} mol$
$\therefore $ moles of $SO _{2}$ in the sample $=3 \times 1.5 \times 10^{-3} mol$
$=$ mol of $FeS$
$\therefore $ Mass of pure $FeS$ in $10$ gram sample $=4.5 \times 10^{-3} \times 88 g$
$\approx 4$
$\therefore \%$ of pure $FeS$ in the sample $=3.96 \%=4 \%$