Đáp án:

 Fe: 7,05%; Cr: 2,48%

Giải thích các bước giải:

+ Sau khi qua ống khử Walden: $F{e^{3 + }} + e \to F{e^{2 + }}$

$5F{e^{2 + }} + MnO_4^ -  + 8{H^ + } \to 5F{e^{3 + }} + M{n^{2 + }} + 4{H_2}O$

$\begin{gathered}
   \Rightarrow {n_{Fe}} = 5{n_{KMn{O_4}}} = 5.0,0192.0,01322 = 1,{26912.10^{ - 3}}mol \hfill \\
   \Rightarrow {m_{Fe}} = 1,{26912.10^{ - 3}}.56 = 0,071g \hfill \\ 
\end{gathered} $

⇒ Trong 10,065 g thép chứa ${m_{Fe}} = 0,071.10 = 0,71g$

$ \Rightarrow \% {m_{Fe}} = \dfrac{{0,71}}{{10,065}}.100\%  = 7,05\% $

+ Sau khi qua ống khử Jone: 

$F{e^{3 + }} + e \to F{e^{2 + }}$

$C{r^{3 + }} + e \to C{r^{2 + }}$

$5F{e^{2 + }} + MnO_4^ -  + 8{H^ + } \to 5F{e^{3 + }} + M{n^{2 + }} + 4{H_2}O$

$5C{r^{2 + }} + MnO_4^ -  + 8{H^ + } \to 5C{r^{3 + }} + M{n^{2 + }} + 4{H_2}O$

${n_{KMn{O_4}}} = 0,03643.0,0192 = {7.10^{ - 4}}$

$\begin{gathered}
   \Rightarrow {n_{Fe}} + {n_{Cr}} = 5{n_{KMn{O_4}}} = 3,{5.10^{ - 3}}mol \hfill \\
   \Rightarrow {n_{Cr}} = 3,{5.10^{ - 3}} - 1,{26912.10^{ - 3}}.2 = 9,{6176.10^{ - 4}}(mol) \hfill \\ 
\end{gathered} $

⇒ Trong 10,065g chứa ${m_{Cr}} = 9,{6176.10^{ - 4}}.52.5 = 0,25g$

$ \Rightarrow \% {m_{Cr}} = \dfrac{{0,25}}{{10,065}}.100\%  = 2,48\% $