Trắc nghiệm Bài 34. Crom và hợp chất của crom (tiếp) - Hóa 12

$\left\{ \begin{gathered}  F{\text{e}}_{2}{\text{O}}_{3} \hfill \\  C{\text{r}}_{2}{\text{O}}_{3} \hfill \\  Al_{2}{\text{O}}_{3} \hfill \\ \end{gathered}  \right.$ $\xrightarrow{+NaOH(\mathsf{dac,du)}}$$\left\{ \begin{align}  & F\text{e}_{2}\text{O}_{3}\downarrow  \\  & C\text{r}_{2}\text{O}_{3},Al{2}\text{O}_{3}:\tan  \\ \end{align} \right.$

=> ${{n}_{F{{\text{e}}_{2}}{{\text{O}}_{\text{3}}}}}\text{ = 0}\text{,1(mol)}$

$\left\{ \begin{gathered}  F{\text{e}}_{2}{\text{O}}_{3} \hfill \\  C{\text{r}}_{2}{\text{O}}_{3} \hfill \\  Al_{2}{\text{O}}_{3} \hfill \\ \end{gathered}  \right.$$\xrightarrow{Al({{t}^{o}})}$$\left\{ \begin{gathered}  C{\text{r}},F{\text{e}} \hfill \\  Al_{2}{\text{O}}_{3} \hfill \\ \end{gathered}  \right.$

Cr₂O₃ + 2Al $\xrightarrow{{{t}^{o}}}$ 2Cr + Al₂O₃

Fe₂O₃ + 2Al $\xrightarrow{{{t}^{o}}}$ 2Fe + Al₂O₃

$=> \frac{1}{2}{{n}_{Al\,\,pu}}={{n}_{C{{\text{r}}_{2}}{{\text{O}}_{3}}}}+{{n}_{F{{\text{e}}_{2}}{{\text{O}}_{3}}}}=\frac{1}{2}.\frac{10,8}{27}=0,2(mol)$

$=> {{n}_{C{{\text{r}}_{2}}{{\text{O}}_{3}}}}=0,2-{{n}_{F{{\text{e}}_{2}}{{\text{O}}_{3}}}}=0,2-0,1=0,1(mol)$

$=>\% {m}_{C{\text{r}}2{\text{O}}3(X)} = \frac{{152.0,1}}{{41,4}}.100\%  = 36,71\% $.

${m}_{Cr}{\text{ }} = \frac{{216.65}}{{100}} = 140,4(kg)$

=>n_Cr = 2,7(kmol)

Bảo toàn nguyên tố Cr;

2n_Cr = n_cromit = 2,7 (kmol)

=> n_{cromit (LT)} = 1,35 (kmol)

=> n_{cromit (TT)} = $\frac{1,35.100}{90}=1,5(kmol)$

=> %quặng = $\frac{1,5.224}{1000}.100%=\text{ }33,6%$

=> %tạp chất = 66,4%