Question

Đã trả lời bởi chuyên gia

Answer 1

Chọn C.

Quá trình:

$X \{\begin{cases} F e; \\ F e (N O_{3})_{2} \\ F e (N O_{3})_{3} \\ F e C O_{3} \end{cases} \overset{t °}{\to} <\begin{array}{l} \underset{H o n hopkhiZ}{\underset{⏟}{N O_{2}, C O_{2}}} ({\bar{M}}_{Z} = 45, 5) \\ Y + \underset{dungdichhonhop}{\underset{⏟}{{\overset{⏞}{N a N O_{3}}}^{0, 04 m o l}, {\overset{⏞}{K H S O_{4}}}^{0, 92 m o l}}} \to \underset{21, 23 g a m}{\underset{⏟}{F e^{n +}; {\overset{⏞}{N a^{+}}}^{0, 04}; {\overset{⏞}{K^{+}}}^{0, 92} : {\overset{⏞}{S O_{4}^{2 -}}}^{0, 92}}} + \underset{honhopkhi}{\underset{⏟}{H_{2}, N O}} ({\bar{M}}_{k h ́ Y} = 13, 2) \end{array}$

Ta có: $m_{F e^{n +}} + m_{S O_{4}^{2 -}} + m_{K^{+}} + m_{N a^{+}} = 143, 04 \to m_{F e^{n +}} = l 7, 92 (g)$

$\overset{B T : N}{\to} n_{N a N O_{3}} = n_{N O} = 0, 04 m o l m à M = \frac{M_{H_{2}} + M_{N O}}{2} = 13, 2 \to n_{H_{2}} = 0, 06 m o l \overset{B T : H}{\to} n_{H_{2} O} = 0, 5 n_{K H S O_{4}} - n_{H_{2}} = 0, 4 m o l \overset{B T : O}{\to} n_{O (Y)} + 3 n_{N a N O_{3}} = n_{N O} + n_{H_{2} O} \to n_{O (Y)} = 0, 32 m o l {\bar{M}}_{X} = \frac{M_{N O_{2}} + M_{C O_{2}}}{2} = 45 \to \{\begin{cases} n_{N O_{2}} = 0, 24 \\ n_{C O_{2}} = 0, 08 \end{cases} \Rightarrow m_{X} = m_{F e} + 62 n_{N O_{3}^{-}} + 60 n_{C O_{3}^{2 -}} = 37, 6 (g)$

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Answer 2

Chọn C.

Quá trình:

$X \{\begin{cases} F e; \\ F e (N O_{3})_{2} \\ F e (N O_{3})_{3} \\ F e C O_{3} \end{cases} \overset{t °}{\to} <\begin{array}{l} \underset{H o n hopkhiZ}{\underset{⏟}{N O_{2}, C O_{2}}} ({\bar{M}}_{Z} = 45, 5) \\ Y + \underset{dungdichhonhop}{\underset{⏟}{{\overset{⏞}{N a N O_{3}}}^{0, 04 m o l}, {\overset{⏞}{K H S O_{4}}}^{0, 92 m o l}}} \to \underset{21, 23 g a m}{\underset{⏟}{F e^{n +}; {\overset{⏞}{N a^{+}}}^{0, 04}; {\overset{⏞}{K^{+}}}^{0, 92} : {\overset{⏞}{S O_{4}^{2 -}}}^{0, 92}}} + \underset{honhopkhi}{\underset{⏟}{H_{2}, N O}} ({\bar{M}}_{k h ́ Y} = 13, 2) \end{array}$

Ta có: $m_{F e^{n +}} + m_{S O_{4}^{2 -}} + m_{K^{+}} + m_{N a^{+}} = 143, 04 \to m_{F e^{n +}} = l 7, 92 (g)$

$\overset{B T : N}{\to} n_{N a N O_{3}} = n_{N O} = 0, 04 m o l m à M = \frac{M_{H_{2}} + M_{N O}}{2} = 13, 2 \to n_{H_{2}} = 0, 06 m o l \overset{B T : H}{\to} n_{H_{2} O} = 0, 5 n_{K H S O_{4}} - n_{H_{2}} = 0, 4 m o l \overset{B T : O}{\to} n_{O (Y)} + 3 n_{N a N O_{3}} = n_{N O} + n_{H_{2} O} \to n_{O (Y)} = 0, 32 m o l {\bar{M}}_{X} = \frac{M_{N O_{2}} + M_{C O_{2}}}{2} = 45 \to \{\begin{cases} n_{N O_{2}} = 0, 24 \\ n_{C O_{2}} = 0, 08 \end{cases} \Rightarrow m_{X} = m_{F e} + 62 n_{N O_{3}^{-}} + 60 n_{C O_{3}^{2 -}} = 37, 6 (g)$

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