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Câu hỏi: Nung 41,48 gam hỗn hợp rắn gồm Fe(NO3)3, Fe(NO3)2 và Cu(NO3)2, sau một thời gian thu được hỗn hợp rắn X và hỗn hợp khí Y có tỉ khối so với He bằng a. Hòa tan hoàn toàn X trong dung dịch HCl loãng, kết thúc phản ứng thu được dung dịch Z chỉ chứa các muối có khối lượng 41,5 gam và 672 ml khí NO (sản phẩm khử duy nhất của N+5; đo đktc). Dung dịch Y hòa tan tối đa 4,48 gam bột Cu. Nếu cho dung dịch NaOH dư vào Y, thấy lượng NaOH phản ứng là 23,2 gam. Giá trị gần nhất của a là
A. 11.
B. 9.
C. 8.
D. 10.
A. 11.
B. 9.
C. 8.
D. 10.
$\text{41,48 gam }\left\langle \begin{aligned}
& Fe{{(N{{O}_{3}})}_{3}} \\
& \text{Fe(N}{{\text{O}}_{3}}{{\text{)}}_{2}} \\
& Cu{{(N{{O}_{3}})}_{2}} \\
\end{aligned} \right\rangle \xrightarrow{{{t}^{0}}}\left\{ \begin{aligned}
& Y\left\langle N{{O}_{2}}\text{ + }{{\text{O}}_{2}} \right\rangle \\
& X\xrightarrow{+HCl}\left\{ \begin{aligned}
& \left\langle NO\text{ (0,03) + }{{\text{H}}_{2}}O \right\rangle \\
& dd\text{ Z}\left\langle \begin{aligned}
& C{{u}^{2+}}\text{; F}{{\text{e}}^{3+}} \\
& C{{l}^{-}}\text{ (x); NO}_{3}^{-}\text{ (y)} \\
\end{aligned} \right\rangle \left\{ \begin{aligned}
& \xrightarrow[\text{0,07}]{+Cu}\left\langle \text{C}{{\text{u}}^{2+}}\text{ + F}{{\text{e}}^{2+}} \right\rangle \\
& \xrightarrow[\text{0,58}]{+NaOH} \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\xrightarrow[Z+Cu]{\text{BT e}}{{\text{n}}_{F{{e}^{3+}}(Z)}}\text{ = 0,14 mol}$
$\xrightarrow{Z+NaOH}\text{ 2}{{\text{n}}_{C{{u}^{2+}}}}\text{ + 3}{{\text{n}}_{F{{e}^{3+}}}}\text{ = }{{\text{n}}_{O{{H}^{-}}}}\xrightarrow{{}}{{\text{n}}_{C{{u}^{2+}}(Z)}}\text{ = 0,08 mol}$
$\Rightarrow \left\{ \begin{aligned}
& \xrightarrow{{{m}_{Z}}}\text{ 35,5x + 62y = 41,5} \\
& \xrightarrow{BTDT.Z}\text{ x + y = 0,14*3 + 0,08*2} \\
\end{aligned} \right.\xrightarrow{{}}\left\{ \begin{aligned}
& \text{x = 0,28 = }{{\text{n}}_{HCl}}\xrightarrow{X+HCl}{{\text{n}}_{{{H}_{2}}O}}\text{ = 0,14 mol} \\
& \text{y = 0,3} \\
\end{aligned} \right.$
$\xrightarrow{41,48}{{\text{m}}_{Fe}}\text{ + }{{\text{m}}_{Cu}}\text{ + }{{\text{m}}_{NO_{3}^{-}}}\xrightarrow{{}}{{\text{n}}_{NO_{3}^{-}(hh\text{})}}\text{ = 0,46 mol}$
$\xrightarrow{BT\text{ N}}\text{ 0,46 = }{{\text{n}}_{N{{O}_{2}}(Y)}}\text{ + }{{\text{n}}_{NO}}\text{ + }{{\text{n}}_{NO_{3}^{-}(Z)}}\xrightarrow{{}}{{\text{n}}_{N{{O}_{2}}(Y)}}\text{ = 0,13 mol}$
Ta thấy: $NO{{_{3}^{-}}_{(hhd)}}\xrightarrow{{}}\left\langle N{{O}_{2}}\text{ + }{{\text{O}}_{2}} \right\rangle \text{ + }\left\langle NO\text{ + }{{\text{H}}_{2}}O \right\rangle \text{ + NO}{{_{3}^{-}}_{(Z)}}$
$\xrightarrow{BT\text{ O}}\text{ 0,46*3 = 0,13*2 + 2}{{\text{n}}_{{{O}_{2}}}}\text{ + 0,03 + 0,14 + 0,3*3 }\xrightarrow{{}}{{\text{n}}_{{{O}_{2}}(Y)}}\text{ = 0,025}$
$\xrightarrow{{}}{{\text{d}}_{Y/He}}\text{ = 10,935}$
& Fe{{(N{{O}_{3}})}_{3}} \\
& \text{Fe(N}{{\text{O}}_{3}}{{\text{)}}_{2}} \\
& Cu{{(N{{O}_{3}})}_{2}} \\
\end{aligned} \right\rangle \xrightarrow{{{t}^{0}}}\left\{ \begin{aligned}
& Y\left\langle N{{O}_{2}}\text{ + }{{\text{O}}_{2}} \right\rangle \\
& X\xrightarrow{+HCl}\left\{ \begin{aligned}
& \left\langle NO\text{ (0,03) + }{{\text{H}}_{2}}O \right\rangle \\
& dd\text{ Z}\left\langle \begin{aligned}
& C{{u}^{2+}}\text{; F}{{\text{e}}^{3+}} \\
& C{{l}^{-}}\text{ (x); NO}_{3}^{-}\text{ (y)} \\
\end{aligned} \right\rangle \left\{ \begin{aligned}
& \xrightarrow[\text{0,07}]{+Cu}\left\langle \text{C}{{\text{u}}^{2+}}\text{ + F}{{\text{e}}^{2+}} \right\rangle \\
& \xrightarrow[\text{0,58}]{+NaOH} \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\xrightarrow[Z+Cu]{\text{BT e}}{{\text{n}}_{F{{e}^{3+}}(Z)}}\text{ = 0,14 mol}$
$\xrightarrow{Z+NaOH}\text{ 2}{{\text{n}}_{C{{u}^{2+}}}}\text{ + 3}{{\text{n}}_{F{{e}^{3+}}}}\text{ = }{{\text{n}}_{O{{H}^{-}}}}\xrightarrow{{}}{{\text{n}}_{C{{u}^{2+}}(Z)}}\text{ = 0,08 mol}$
$\Rightarrow \left\{ \begin{aligned}
& \xrightarrow{{{m}_{Z}}}\text{ 35,5x + 62y = 41,5} \\
& \xrightarrow{BTDT.Z}\text{ x + y = 0,14*3 + 0,08*2} \\
\end{aligned} \right.\xrightarrow{{}}\left\{ \begin{aligned}
& \text{x = 0,28 = }{{\text{n}}_{HCl}}\xrightarrow{X+HCl}{{\text{n}}_{{{H}_{2}}O}}\text{ = 0,14 mol} \\
& \text{y = 0,3} \\
\end{aligned} \right.$
$\xrightarrow{41,48}{{\text{m}}_{Fe}}\text{ + }{{\text{m}}_{Cu}}\text{ + }{{\text{m}}_{NO_{3}^{-}}}\xrightarrow{{}}{{\text{n}}_{NO_{3}^{-}(hh\text{})}}\text{ = 0,46 mol}$
$\xrightarrow{BT\text{ N}}\text{ 0,46 = }{{\text{n}}_{N{{O}_{2}}(Y)}}\text{ + }{{\text{n}}_{NO}}\text{ + }{{\text{n}}_{NO_{3}^{-}(Z)}}\xrightarrow{{}}{{\text{n}}_{N{{O}_{2}}(Y)}}\text{ = 0,13 mol}$
Ta thấy: $NO{{_{3}^{-}}_{(hhd)}}\xrightarrow{{}}\left\langle N{{O}_{2}}\text{ + }{{\text{O}}_{2}} \right\rangle \text{ + }\left\langle NO\text{ + }{{\text{H}}_{2}}O \right\rangle \text{ + NO}{{_{3}^{-}}_{(Z)}}$
$\xrightarrow{BT\text{ O}}\text{ 0,46*3 = 0,13*2 + 2}{{\text{n}}_{{{O}_{2}}}}\text{ + 0,03 + 0,14 + 0,3*3 }\xrightarrow{{}}{{\text{n}}_{{{O}_{2}}(Y)}}\text{ = 0,025}$
$\xrightarrow{{}}{{\text{d}}_{Y/He}}\text{ = 10,935}$
Đáp án A.