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Câu hỏi: Nung hỗn hợp X gồm Al, Fe và Cu (trong đó Cu chiếm 34,72% khối lượng) trong không khí một thời gian, thu được 6,17 gam hỗn hợp rắn Y gồm các kim loại và oxit tương ứng. Cho hỗn hợp Y tác dụng với dung dịch A chứa 0,36 mol $KHS{{O}_{4}}$ và 0,04 mol $KN{{O}_{3}}$. Sau phản ứng, thu được dung dịch B chỉ chứa 56,05 gam muối sunfat trung hòa (không làm mất màu thuốc tím) và thoát ra 336 ml hỗn hợp khí Z chứa các hợp chất của nitơ có ${{d}_{Z/{{H}_{2}}}}=20$. Cho dung dịch B tác dụng hoàn toàn với 170 ml dung dịch NaOH 2M thì thu được m gam kết tủa. Giá trị của m gần nhất với giá trị nào sau đây?
A. 11,3 gam.
B. 8,3 gam.
C. 9,6 gam.
D. 8,9 gam.
A. 11,3 gam.
B. 8,3 gam.
C. 9,6 gam.
D. 8,9 gam.
$X\left\{ \begin{aligned}
& Al \\
& Fe \\
& Cu \\
\end{aligned} \right.\xrightarrow[KK]{t{}^\circ }\underbrace{\text{rắn Y}\left. \left\langle \begin{aligned}
& Oxit\text{ KL} \\
& \text{KL dư}\!\! \\
\end{aligned} \right. \right|}_{6,17\left( g \right)}+A\left\{ \begin{aligned}
& KHS{{O}_{4}}:0,36\left( mol \right) \\
& KN{{O}_{3}}:0,04\left( mol \right) \\
\end{aligned} \right.\left\langle \begin{aligned}
& \underbrace{B\left\{ \begin{aligned}
& C{{u}^{2+}} \\
& F{{e}^{3+}} \\
& A{{l}^{3+}} \\
& NH_{4}^{+}:0,02\left( mol \right) \\
& {{K}^{+}}:0,4\left( mol \right) \\
& SO_{4}^{2-} \\
\end{aligned} \right.}_{56,05\left( g \right)} \\
& \underbrace{\text{Kh }\!\!\acute{\mathrm{i}}\!\!\text{ Z}\left( Oxit\text{ Nitơ}\!\! \right)}_{0,6\left( gam \right)}:0,015\left( mol \right) \\
\end{aligned} \right.+\underbrace{{{H}_{2}}O}_{2,52\left( gam \right)}$
$BTKL:{{m}_{Y}}+{{m}_{A}}={{m}_{B}}+{{m}_{Z}}+{{m}_{{{H}_{2}}O}}\to {{m}_{{{H}_{2}}O}}=2,52\left( gam \right)\to {{n}_{{{H}_{2}}O}}=0,14\left( mol \right)$
$BTH:{{n}_{KHS{{O}_{4}}}}=4{{n}_{NH_{4}^{+}}}+2{{n}_{{{H}_{2}}O}}\to {{n}_{NH_{4}^{+}}}=\dfrac{0,36-2.0,14}{4}=0,02\left( mol \right)$
$BTN:{{n}_{N\left( Z \right)}}={{n}_{KN{{O}_{3}}}}-{{n}_{NH_{4}^{+}}}=0,02\left( mol \right);{{m}_{Z}}={{m}_{N\left( Z \right)}}+{{m}_{O\left( Z \right)}}=0,6\left( gam \right)$
$\to {{m}_{O\left( Z \right)}}=0,32\left( gam \right)\to {{n}_{O\left( Z \right)}}=0,02\left( mol \right)$
$BTO:{{n}_{O\left( Y \right)}}+3{{n}_{KN{{O}_{3}}}}={{n}_{O\left( Z \right)}}+{{n}_{{{H}_{2}}O}}\to {{n}_{O\left( Y \right)}}=0,04\left( mol \right)$
${{m}_{KL\left( X \right)}}={{m}_{Y}}-{{m}_{O\left( Y \right)}}=6,17-0,04.16=5,53\left( gam \right)$ (1)
Do Cu chiếm 34,72% về khối lượng trong X $\to {{m}_{Cu}}=1,92\left( gam \right)\to {{n}_{Cu}}=0,03\left( mol \right)$
$BTe:{{n}_{e\left( KL \right)}}=2{{n}_{SO_{4}^{2-}}}-{{n}_{{{K}^{+}}}}-{{n}_{NH_{4}^{+}}}=2.0,36-0,4-0,02=0,3\left( mol \right)$ (2)
Từ (1), (2): $\left\{ \begin{aligned}
& 56a+27b+\underbrace{1,92}_{{{m}_{Ca}}}=5,53 \\
& 3a+3b+\underbrace{0,03.2}_{{{n}_{e\left( Cu \right)}}}=0,3 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,05={{n}_{Fe}} \\
& b=0,03={{n}_{Al}} \\
\end{aligned} \right.$
$B+NaOH$ lượng $O{{H}^{-}}$ dùng để kết hợp các cation $\left( C{{u}^{2+}},F{{e}^{3+}},A{{l}^{3+}},NH_{4}^{+} \right)$ là:
${{n}_{O{{H}^{-}}}}=\underbrace{0,03.3}_{{{n}_{A{{l}^{3+}}}}}+\underbrace{0,05.3}_{{{n}_{F{{e}^{3+}}}}}+\underbrace{0,03.2}_{{{n}_{C{{u}^{2+}}}}}+\underbrace{0,02}_{{{n}_{NH_{4}^{+}}}}=0,32\left( mol \right)\to {{n}_{O{{H}^{-}}\text{dư}\!\!}}=\underbrace{0,34}_{\text{ban đầu}}-\underbrace{0,32}_{\left( \text{pứ}\!\! \right)}=0,02\left( mol \right)$
$0,02 mol OH_{\text{dư}\!\!}^{-}$ hòa tan $Al{{\left( OH \right)}_{3}}\to Al\left( OH \right)_{4}^{-}\to {{n}_{Al{{\left( OH \right)}_{3}}\text{ dư}\!\!}}=0,03-0,02=0,01\left( mol \right)$
$\Rightarrow {{m}_{\text{kết tủa}}}=\underbrace{{{m}_{Fe{{\left( OH \right)}_{3}}}}}_{0,05.107}+\underbrace{{{m}_{Cu{{\left( OH \right)}_{2}}}}}_{0,03.98}+\underbrace{{{m}_{Al{{\left( OH \right)}_{3\left( \text{dư}\!\! \right)}}}}}_{0,01.78}=9,07\left( gam \right)$.
& Al \\
& Fe \\
& Cu \\
\end{aligned} \right.\xrightarrow[KK]{t{}^\circ }\underbrace{\text{rắn Y}\left. \left\langle \begin{aligned}
& Oxit\text{ KL} \\
& \text{KL dư}\!\! \\
\end{aligned} \right. \right|}_{6,17\left( g \right)}+A\left\{ \begin{aligned}
& KHS{{O}_{4}}:0,36\left( mol \right) \\
& KN{{O}_{3}}:0,04\left( mol \right) \\
\end{aligned} \right.\left\langle \begin{aligned}
& \underbrace{B\left\{ \begin{aligned}
& C{{u}^{2+}} \\
& F{{e}^{3+}} \\
& A{{l}^{3+}} \\
& NH_{4}^{+}:0,02\left( mol \right) \\
& {{K}^{+}}:0,4\left( mol \right) \\
& SO_{4}^{2-} \\
\end{aligned} \right.}_{56,05\left( g \right)} \\
& \underbrace{\text{Kh }\!\!\acute{\mathrm{i}}\!\!\text{ Z}\left( Oxit\text{ Nitơ}\!\! \right)}_{0,6\left( gam \right)}:0,015\left( mol \right) \\
\end{aligned} \right.+\underbrace{{{H}_{2}}O}_{2,52\left( gam \right)}$
$BTKL:{{m}_{Y}}+{{m}_{A}}={{m}_{B}}+{{m}_{Z}}+{{m}_{{{H}_{2}}O}}\to {{m}_{{{H}_{2}}O}}=2,52\left( gam \right)\to {{n}_{{{H}_{2}}O}}=0,14\left( mol \right)$
$BTH:{{n}_{KHS{{O}_{4}}}}=4{{n}_{NH_{4}^{+}}}+2{{n}_{{{H}_{2}}O}}\to {{n}_{NH_{4}^{+}}}=\dfrac{0,36-2.0,14}{4}=0,02\left( mol \right)$
$BTN:{{n}_{N\left( Z \right)}}={{n}_{KN{{O}_{3}}}}-{{n}_{NH_{4}^{+}}}=0,02\left( mol \right);{{m}_{Z}}={{m}_{N\left( Z \right)}}+{{m}_{O\left( Z \right)}}=0,6\left( gam \right)$
$\to {{m}_{O\left( Z \right)}}=0,32\left( gam \right)\to {{n}_{O\left( Z \right)}}=0,02\left( mol \right)$
$BTO:{{n}_{O\left( Y \right)}}+3{{n}_{KN{{O}_{3}}}}={{n}_{O\left( Z \right)}}+{{n}_{{{H}_{2}}O}}\to {{n}_{O\left( Y \right)}}=0,04\left( mol \right)$
${{m}_{KL\left( X \right)}}={{m}_{Y}}-{{m}_{O\left( Y \right)}}=6,17-0,04.16=5,53\left( gam \right)$ (1)
Do Cu chiếm 34,72% về khối lượng trong X $\to {{m}_{Cu}}=1,92\left( gam \right)\to {{n}_{Cu}}=0,03\left( mol \right)$
$BTe:{{n}_{e\left( KL \right)}}=2{{n}_{SO_{4}^{2-}}}-{{n}_{{{K}^{+}}}}-{{n}_{NH_{4}^{+}}}=2.0,36-0,4-0,02=0,3\left( mol \right)$ (2)
Từ (1), (2): $\left\{ \begin{aligned}
& 56a+27b+\underbrace{1,92}_{{{m}_{Ca}}}=5,53 \\
& 3a+3b+\underbrace{0,03.2}_{{{n}_{e\left( Cu \right)}}}=0,3 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,05={{n}_{Fe}} \\
& b=0,03={{n}_{Al}} \\
\end{aligned} \right.$
$B+NaOH$ lượng $O{{H}^{-}}$ dùng để kết hợp các cation $\left( C{{u}^{2+}},F{{e}^{3+}},A{{l}^{3+}},NH_{4}^{+} \right)$ là:
${{n}_{O{{H}^{-}}}}=\underbrace{0,03.3}_{{{n}_{A{{l}^{3+}}}}}+\underbrace{0,05.3}_{{{n}_{F{{e}^{3+}}}}}+\underbrace{0,03.2}_{{{n}_{C{{u}^{2+}}}}}+\underbrace{0,02}_{{{n}_{NH_{4}^{+}}}}=0,32\left( mol \right)\to {{n}_{O{{H}^{-}}\text{dư}\!\!}}=\underbrace{0,34}_{\text{ban đầu}}-\underbrace{0,32}_{\left( \text{pứ}\!\! \right)}=0,02\left( mol \right)$
$0,02 mol OH_{\text{dư}\!\!}^{-}$ hòa tan $Al{{\left( OH \right)}_{3}}\to Al\left( OH \right)_{4}^{-}\to {{n}_{Al{{\left( OH \right)}_{3}}\text{ dư}\!\!}}=0,03-0,02=0,01\left( mol \right)$
$\Rightarrow {{m}_{\text{kết tủa}}}=\underbrace{{{m}_{Fe{{\left( OH \right)}_{3}}}}}_{0,05.107}+\underbrace{{{m}_{Cu{{\left( OH \right)}_{2}}}}}_{0,03.98}+\underbrace{{{m}_{Al{{\left( OH \right)}_{3\left( \text{dư}\!\! \right)}}}}}_{0,01.78}=9,07\left( gam \right)$.
Đáp án D.