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\% $
