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 $(y)$ = $(1.05 \ g / 5.6 \ g) 	imes 100 = 18.75 \%$.

List-$I$	List-$II$
$P. \ PbO_2 + H_2SO_4 \xrightarrow{?} PbSO_4 + O_2 + \text{other product}$	$1. \ NO$
$Q. \ Na_2S_2O_3 + H_2O \xrightarrow{?} NaHSO_4 + \text{other product}$	$2. \ I_2$
$R. \ N_2H_4 \xrightarrow{?} N_2 + \text{other product}$	$3. \ \text{Warm}$
$S. \ XeF_2 \xrightarrow{?} Xe + \text{other product}$	$4. \ Cl_2$

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