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Câu hỏi: Cho dãy biến đổi sau: $Cr\xrightarrow{+HCl}X\xrightarrow{+C{{l}_{2}}}Y\xrightarrow{+NaOH\text{ d}}Z\xrightarrow{+B{{r}_{2}}+ddNaOH}T$
X, Y, Z, T là
A. CrCl2, CrCl3, NaCrO2, Na2Cr2O7
B. CrCl2, CrCl3, Cr(OH)3, Na2CrO4
C. CrCl2, CrCl3, NaCrO2, Na2CrO4
D. CrCl2, CrCl3, Cr(OH)3, Na2CrO7
X, Y, Z, T là
A. CrCl2, CrCl3, NaCrO2, Na2Cr2O7
B. CrCl2, CrCl3, Cr(OH)3, Na2CrO4
C. CrCl2, CrCl3, NaCrO2, Na2CrO4
D. CrCl2, CrCl3, Cr(OH)3, Na2CrO7
$\begin{aligned}
& Cr+2HCl\xrightarrow{{}}CrC{{l}_{2}}+{{H}_{2}}\uparrow \\
& CrC{{l}_{2}}+\dfrac{1}{2}C{{l}_{2}}\xrightarrow[{}]{}CrC{{l}_{3}} \\
& CrC{{l}_{3}}+4NaO{{H}_{\left( \text{d} \right)}}\xrightarrow[{}]{}NaCr{{O}_{2}}+3NaCl+2{{H}_{2}}O \\
& 2NaCr{{O}_{2}}+3B{{r}_{2}}+8NaOH\xrightarrow{{}}2N{{a}_{2}}Cr{{O}_{4}}+6NaBr+4{{H}_{2}}O \\
\end{aligned}$
& Cr+2HCl\xrightarrow{{}}CrC{{l}_{2}}+{{H}_{2}}\uparrow \\
& CrC{{l}_{2}}+\dfrac{1}{2}C{{l}_{2}}\xrightarrow[{}]{}CrC{{l}_{3}} \\
& CrC{{l}_{3}}+4NaO{{H}_{\left( \text{d} \right)}}\xrightarrow[{}]{}NaCr{{O}_{2}}+3NaCl+2{{H}_{2}}O \\
& 2NaCr{{O}_{2}}+3B{{r}_{2}}+8NaOH\xrightarrow{{}}2N{{a}_{2}}Cr{{O}_{4}}+6NaBr+4{{H}_{2}}O \\
\end{aligned}$
Đáp án C.