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Câu hỏi: Hòa tan hết 15,0 gam hỗn hợp $X$ gồm $Fe,F{{e}_{3}}{{O}_{4}},FeC{{O}_{3}}$ và $Fe{{\left( N{{O}_{3}} \right)}_{2}}$ trong dung dịch chứa $NaHS{{O}_{4}}$ và 0,16 mol $HN{{O}_{3}}$, thu được dung dịch $Y$ và hỗn hợp khí $Z$ gồm $C{{O}_{2}}$ và $NO$ (tỉ lệ mol tương ứng $1:4$ ). Dung dịch $Y$ hòa tan tối đa 8,64 gam bột $Cu$, thấy thoát ra 0,03 mol khí $NO$. Nếu cho dung dịch $Ba{{\left( OH \right)}_{2}}$ dư vào $Y$, thu được 154,4 gam kết tủa. Biết các phản ứng xảy ra hoàn toàn và khí $NO$ là sản phẩm khử duy nhất của cả quá trình. Phần trăm khối lượng của $Fe$ đơn chất trong hỗn hợp $X$ là:
A. 48,80%
B. 33,60%
C. 37,33%
D. 29,87%
A. 48,80%
B. 33,60%
C. 37,33%
D. 29,87%
$\begin{aligned}
& X\left\{ \begin{aligned}
& Fe \\
& F{{e}_{3}}{{O}_{4}} \\
& FeC{{O}_{3}} \\
& \underbrace{Fe{{\left( N{{O}_{3}} \right)}_{2}}}_{15 gam} \\
\end{aligned} \right.+\left\{ \begin{aligned}
& NaHS{{O}_{4}}:0,58 \\
& HN{{O}_{3}}:0,16 \\
\end{aligned} \right.\xrightarrow[TN1]{}\left| \begin{aligned}
& Y\left\{ \begin{aligned}
& F{{e}^{3+}} \\
& N{{a}^{+}}:0,58 \\
& SO_{4}^{2-}:0,58 \\
& NO_{3}^{-}\left( du \right) \\
& H_{du}^{+}=4{{n}_{NO}}=0,12 \\
\end{aligned} \right.+\left| \begin{aligned}
& \xrightarrow{+Cu:0,135\ \left( TN2 \right)}\left\{ \begin{aligned}
& NO\uparrow :0,03 \\
& F{{e}^{3+}} \\
\end{aligned} \right. \\
& \xrightarrow{+Ba{{\left( OH \right)}_{2}}\ \left( TN2 \right)}\underbrace{\downarrow \left\{ \begin{aligned}
& Fe{{\left( OH \right)}_{3}}:0,18 \\
& BaS{{O}_{4}}:0,58 \\
\end{aligned} \right.}_{154,4\ gam} \\
\end{aligned} \right. \\
& Z\left\{ \begin{aligned}
& C{{O}_{2}}:a \\
& NO:4a \\
\end{aligned} \right. \\
& {{H}_{2}}O:\dfrac{0,58+0,16-0,12}{3}=0,31\ mol\left( BT.H \right) \\
\end{aligned} \right. \\
& \xrightarrow{BTE\ cho\ TN2}2{{n}_{Cu}}={{n}_{F{{e}^{3+}}\left( Y \right)}}+3{{n}_{NO}}\to {{n}_{F{{e}^{3+}}\left( Y \right)}}=2.0,135-3.0,03=0,18\ mol \\
& \xrightarrow{BTT\ cho\ Y}{{n}_{NO_{3}^{-}\left( Y \right)}}=3{{n}_{F{{e}^{3+}}}}+{{n}_{N{{a}^{+}}}}+{{n}_{{{H}^{+}}}}-2{{n}_{SO_{4}^{2-}}}=3.0,18+0,58+0,12-2.0,58=0,08\ mol \\
& {{m}_{Y}}=\left( 56.0,18+23.0,58+96.0,58+62.0,08+0,12 \right)=84,18\ gam \\
& \xrightarrow{BTKL\ cho\ TN1}15+\left( 120.0,58+63.0,16 \right)=84,18+\left( 44a+30.4a \right)+18.0,31\to a=0,03\ mol \\
& \to \left\{ \begin{aligned}
& \xrightarrow{BT.C}{{n}_{FeC{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,03\ mol \\
& \xrightarrow{BT.N}{{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}=\dfrac{{{n}_{NO_{3}^{-}\left( du \right)}}+{{n}_{NO}}-{{n}_{HN{{O}_{3}}}}}{2}=\dfrac{0,08+4.0,03-0,16}{2}=0,02\ mol \\
& \xrightarrow{BT.O}4{{n}_{F{{e}_{3}}{{O}_{4}}}}+3{{n}_{FeC{{O}_{3}}}}+6{{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}+3{{n}_{HN{{O}_{3}}}}=3{{n}_{NO_{3}^{-}\left( du \right)}}+2{{n}_{C{{O}_{2}}}}+{{n}_{NO}}+{{n}_{{{H}_{2}}O}} \\
& \to {{n}_{F{{e}_{3}}{{O}_{4}}}}=0,01\ mol \\
\end{aligned} \right. \\
& \xrightarrow{BTKL\ cho\ TN1}{{m}_{Fe}}=15-116.0,03-180.0,02-232.0,01=5,6\ gam \\
& \to \%{{m}_{Fe}}=\dfrac{5,6}{15}.100\%=37,33\% \\
\end{aligned}$
& X\left\{ \begin{aligned}
& Fe \\
& F{{e}_{3}}{{O}_{4}} \\
& FeC{{O}_{3}} \\
& \underbrace{Fe{{\left( N{{O}_{3}} \right)}_{2}}}_{15 gam} \\
\end{aligned} \right.+\left\{ \begin{aligned}
& NaHS{{O}_{4}}:0,58 \\
& HN{{O}_{3}}:0,16 \\
\end{aligned} \right.\xrightarrow[TN1]{}\left| \begin{aligned}
& Y\left\{ \begin{aligned}
& F{{e}^{3+}} \\
& N{{a}^{+}}:0,58 \\
& SO_{4}^{2-}:0,58 \\
& NO_{3}^{-}\left( du \right) \\
& H_{du}^{+}=4{{n}_{NO}}=0,12 \\
\end{aligned} \right.+\left| \begin{aligned}
& \xrightarrow{+Cu:0,135\ \left( TN2 \right)}\left\{ \begin{aligned}
& NO\uparrow :0,03 \\
& F{{e}^{3+}} \\
\end{aligned} \right. \\
& \xrightarrow{+Ba{{\left( OH \right)}_{2}}\ \left( TN2 \right)}\underbrace{\downarrow \left\{ \begin{aligned}
& Fe{{\left( OH \right)}_{3}}:0,18 \\
& BaS{{O}_{4}}:0,58 \\
\end{aligned} \right.}_{154,4\ gam} \\
\end{aligned} \right. \\
& Z\left\{ \begin{aligned}
& C{{O}_{2}}:a \\
& NO:4a \\
\end{aligned} \right. \\
& {{H}_{2}}O:\dfrac{0,58+0,16-0,12}{3}=0,31\ mol\left( BT.H \right) \\
\end{aligned} \right. \\
& \xrightarrow{BTE\ cho\ TN2}2{{n}_{Cu}}={{n}_{F{{e}^{3+}}\left( Y \right)}}+3{{n}_{NO}}\to {{n}_{F{{e}^{3+}}\left( Y \right)}}=2.0,135-3.0,03=0,18\ mol \\
& \xrightarrow{BTT\ cho\ Y}{{n}_{NO_{3}^{-}\left( Y \right)}}=3{{n}_{F{{e}^{3+}}}}+{{n}_{N{{a}^{+}}}}+{{n}_{{{H}^{+}}}}-2{{n}_{SO_{4}^{2-}}}=3.0,18+0,58+0,12-2.0,58=0,08\ mol \\
& {{m}_{Y}}=\left( 56.0,18+23.0,58+96.0,58+62.0,08+0,12 \right)=84,18\ gam \\
& \xrightarrow{BTKL\ cho\ TN1}15+\left( 120.0,58+63.0,16 \right)=84,18+\left( 44a+30.4a \right)+18.0,31\to a=0,03\ mol \\
& \to \left\{ \begin{aligned}
& \xrightarrow{BT.C}{{n}_{FeC{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,03\ mol \\
& \xrightarrow{BT.N}{{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}=\dfrac{{{n}_{NO_{3}^{-}\left( du \right)}}+{{n}_{NO}}-{{n}_{HN{{O}_{3}}}}}{2}=\dfrac{0,08+4.0,03-0,16}{2}=0,02\ mol \\
& \xrightarrow{BT.O}4{{n}_{F{{e}_{3}}{{O}_{4}}}}+3{{n}_{FeC{{O}_{3}}}}+6{{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}+3{{n}_{HN{{O}_{3}}}}=3{{n}_{NO_{3}^{-}\left( du \right)}}+2{{n}_{C{{O}_{2}}}}+{{n}_{NO}}+{{n}_{{{H}_{2}}O}} \\
& \to {{n}_{F{{e}_{3}}{{O}_{4}}}}=0,01\ mol \\
\end{aligned} \right. \\
& \xrightarrow{BTKL\ cho\ TN1}{{m}_{Fe}}=15-116.0,03-180.0,02-232.0,01=5,6\ gam \\
& \to \%{{m}_{Fe}}=\dfrac{5,6}{15}.100\%=37,33\% \\
\end{aligned}$
Đáp án C.