A sample (5.6 g) containing iron is complete | vedclass

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Question

(A) The balanced redox reaction is: $5Fe^{2+} + MnO_4^{-} + 8H^{+} \longrightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$.In $25.0 \ mL$ of the solution,the mole

Vedclass · Accepted Answer

$1.87, 18.75$

Vedclass · Answer

(A) The balanced redox reaction is: $5Fe^{2+} + MnO_4^{-} + 8H^{+} \longrightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$.In $25.0 \ mL$ of the solution,the moles of $MnO_4^{-}$ used are: $n(MnO_4^{-}) = M 	imes V = 0.03 \ mol \ L^{-1} 	imes 0.0125 \ L = 3.75 	imes 10^{-4} \ mol$.From the stoichiometry,$1 \ mol$ of $MnO_4^{-}$ reacts with $5 \ mol$ of $Fe^{2+}$.So,$n(Fe^{2+}) 	ext{ in } 25 \ mL = 5 	imes 3.75 	imes 10^{-4} = 1.875 	imes 10^{-3} \ mol$.In $250 \ mL$ of the solution,the total moles of $Fe^{2+}$ are: $1.875 	imes 10^{-3} 	imes 10 = 1.875 	imes 10^{-2} \ mol$.Comparing this with $x 	imes 10^{-2}$,we get $x = 1.875 \approx 1.87$.Mass of iron = $1.875 	imes 10^{-2} \ mol 	imes 56 \ g \ mol^{-1} = 1.05 \ g$.Percentage of iron $(y)$ = $(1.05 \ g / 5.6 \ g) 	imes 100 = 18.75 \%$.

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