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Câu hỏi: Thực hiện phản ứng nhiệt nhôm m gam hỗn hợp Al, CuO, Fe3O4 và Fe2O3 trong khí trơ, thu được hỗn hợp chất rắn X. Cho X vào dung dịch NaOH dư thu được dung dịch Y, chất rắn không tan Z và 0,672 lít H2 (đktc). Sục khí CO2 dư vào Y thu đươc 7,8 gam kết tủa. Cho Z tan hết vào dung dịch H2SO4, thu được dung dịch chứa 16,2 gam muối sunfat và 2,464 lít khí SO2 (đktc, là sản phẩm khử duy nhất của H2SO4). Biết rằng các phản ứng xảy ra hoàn toàn. Giá trị m gần nhất với
A. 10,259.
B. 11,245.
C. 14,289.
D. 12,339.
A. 10,259.
B. 11,245.
C. 14,289.
D. 12,339.
$\left\{ \begin{aligned}
& \text{Al} \\
& {{\text{R}}_{x}}{{\text{O}}_{y}} \\
\end{aligned} \right.\xrightarrow{{{t}^{0}}}\text{ X}\left[ \begin{aligned}
& A{{l}_{2}}{{O}_{3}} \\
& Al\text{ d} \\
& \text{R} \\
\end{aligned} \right]\text{; X }\xrightarrow{NaOH\text{ d}}\left\{ \begin{aligned}
& \text{Z }\left\langle \text{R} \right\rangle \xrightarrow{+{{H}_{2}}S{{O}_{4}}}\text{ 16,2 gam }\left\langle \text{R + SO}_{4}^{2-} \right\rangle \text{ + }\underbrace{\text{S}{{\text{O}}_{2}}}_{\text{0,11}} \\
& \text{dd Y }\xrightarrow{+C{{O}_{2}}}\text{ Al(OH}{{\text{)}}_{3}}\text{ (0,1)} \\
& {{\text{H}}_{2}}\text{ (0,03)} \\
\end{aligned} \right.$
$\xrightarrow[X+NaOH]{\text{BT e}}\text{ 3}{{\text{n}}_{Al(\text{d})}}\text{ = 2}{{\text{n}}_{{{H}_{2}}}}\Rightarrow {{\text{n}}_{Al(\text{d})}}\text{ = 0,02 mol}$
$\xrightarrow{\text{BT Al}}\text{ 2}{{\text{n}}_{A{{l}_{2}}{{O}_{3}}(X)}}\text{ + }{{\text{n}}_{Al(d)}}\text{ = }{{\text{n}}_{Al{{(OH)}_{3}}}}\text{ = }{{\text{n}}_{Al(\text{b})}}\Rightarrow {{\text{n}}_{A{{l}_{2}}{{O}_{3}}(X)}}\text{ = 0,04 mol}$
$\xrightarrow{R+{{H}_{2}}S{{O}_{4}}}{{\text{n}}_{SO_{4}^{2-}(\text{Mu }\!\!\grave{\mathrm{e}}\!\!\text{ i})}}\text{ = }{{\text{n}}_{S{{O}_{2}}}}\Rightarrow {{\text{m}}_{R}}\text{ = }{{\text{m}}_{M}}\text{ - }{{\text{m}}_{SO_{4}^{2-}}}\text{ = 5,64 gam}$
$\xrightarrow{BT\text{ O}}{{\text{n}}_{O(oxit)}}\text{ = 3}{{\text{n}}_{A{{l}_{2}}{{O}_{3}}(X)}}\text{ = 0,12 mol }\xrightarrow{{}}{{\text{m}}_{Oxit}}\text{ = }{{\text{m}}_{R}}\text{ + }{{\text{m}}_{O}}\text{ = 7,56 gam}$
$\xrightarrow{{}}\text{ m = }{{\text{m}}_{Al}}\text{ + }{{\text{m}}_{{{R}_{x}}{{O}_{y}}}}\text{ = 0,1}\text{.27 + 7,56 = 10,26 gam}$
& \text{Al} \\
& {{\text{R}}_{x}}{{\text{O}}_{y}} \\
\end{aligned} \right.\xrightarrow{{{t}^{0}}}\text{ X}\left[ \begin{aligned}
& A{{l}_{2}}{{O}_{3}} \\
& Al\text{ d} \\
& \text{R} \\
\end{aligned} \right]\text{; X }\xrightarrow{NaOH\text{ d}}\left\{ \begin{aligned}
& \text{Z }\left\langle \text{R} \right\rangle \xrightarrow{+{{H}_{2}}S{{O}_{4}}}\text{ 16,2 gam }\left\langle \text{R + SO}_{4}^{2-} \right\rangle \text{ + }\underbrace{\text{S}{{\text{O}}_{2}}}_{\text{0,11}} \\
& \text{dd Y }\xrightarrow{+C{{O}_{2}}}\text{ Al(OH}{{\text{)}}_{3}}\text{ (0,1)} \\
& {{\text{H}}_{2}}\text{ (0,03)} \\
\end{aligned} \right.$
$\xrightarrow[X+NaOH]{\text{BT e}}\text{ 3}{{\text{n}}_{Al(\text{d})}}\text{ = 2}{{\text{n}}_{{{H}_{2}}}}\Rightarrow {{\text{n}}_{Al(\text{d})}}\text{ = 0,02 mol}$
$\xrightarrow{\text{BT Al}}\text{ 2}{{\text{n}}_{A{{l}_{2}}{{O}_{3}}(X)}}\text{ + }{{\text{n}}_{Al(d)}}\text{ = }{{\text{n}}_{Al{{(OH)}_{3}}}}\text{ = }{{\text{n}}_{Al(\text{b})}}\Rightarrow {{\text{n}}_{A{{l}_{2}}{{O}_{3}}(X)}}\text{ = 0,04 mol}$
$\xrightarrow{R+{{H}_{2}}S{{O}_{4}}}{{\text{n}}_{SO_{4}^{2-}(\text{Mu }\!\!\grave{\mathrm{e}}\!\!\text{ i})}}\text{ = }{{\text{n}}_{S{{O}_{2}}}}\Rightarrow {{\text{m}}_{R}}\text{ = }{{\text{m}}_{M}}\text{ - }{{\text{m}}_{SO_{4}^{2-}}}\text{ = 5,64 gam}$
$\xrightarrow{BT\text{ O}}{{\text{n}}_{O(oxit)}}\text{ = 3}{{\text{n}}_{A{{l}_{2}}{{O}_{3}}(X)}}\text{ = 0,12 mol }\xrightarrow{{}}{{\text{m}}_{Oxit}}\text{ = }{{\text{m}}_{R}}\text{ + }{{\text{m}}_{O}}\text{ = 7,56 gam}$
$\xrightarrow{{}}\text{ m = }{{\text{m}}_{Al}}\text{ + }{{\text{m}}_{{{R}_{x}}{{O}_{y}}}}\text{ = 0,1}\text{.27 + 7,56 = 10,26 gam}$
Đáp án A.