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Câu hỏi: Hòa tan hết 31,36 gam hỗn hợp rắn X gồm Mg, Fe, Fe3O4 và FeCO3 vào dung dịch chứa H2SO4 và NaNO3, thu được 4,48 lít (đktc) hỗn hợp khí Y (gồm CO2, NO, N2, H2) có khối lượng 5,14 gam và dung dịch Z chỉ chứa các muối trung hòa. Dung dịch Z phản ứng tối đa với 1,285 mol NaOH, thu được 46,54 gam kết tủa và 0,56 lít khí (đktc). Nếu cho Z tác dụng với dung dịch BaCl2 dư thì thu được 166,595 gam kết tủa. Biết các phản ứng xảy ra hoàn toàn. Phần trăm khối lượng Fe3O4 trong X là
A. 29,59%
B. 36,99%
C. 44,39%
D. 14,80%
A. 29,59%
B. 36,99%
C. 44,39%
D. 14,80%
$\underbrace{X\left\{ \begin{aligned}
& Mg \\
& Fe \\
& F{{e}_{3}}{{O}_{4}} \\
& FeC{{O}_{3}} \\
\end{aligned} \right.}_{31,36gam}+A\left\{ \begin{aligned}
& {{H}_{2}}S{{O}_{4}}:x \\
& NaN{{O}_{3}}:y \\
\end{aligned} \right.\to \left| \begin{aligned}
& \\
& Z\left\{ \begin{aligned}
& N{{a}^{+}} \\
& M{{g}^{2+}} \\
& F{{e}^{2+}} \\
& F{{e}^{3+}} \\
& NH_{4}^{+} \\
& SO_{4}^{2-} \\
\end{aligned} \right.\left| \begin{aligned}
& \xrightarrow[1,285mol]{+NaOH}\left| \begin{aligned}
& 46,54gam\downarrow \left\{ \begin{aligned}
& Mg{{(OH)}_{2}} \\
& Fe{{(OH)}_{2}} \\
& Fe{{(OH)}_{3}} \\
\end{aligned} \right. \\
& N{{a}_{2}}S{{O}_{4}}:(0,5y+0,6425)(BT.Na) \\
& \uparrow N{{H}_{3}}:0,025mol \\
& {{H}_{2}}O \\
\end{aligned} \right. \\
& \xrightarrow{+BaC{{l}_{2}}}\downarrow BaS{{O}_{4}}:x=0,715 \\
\end{aligned} \right. \\
& 5,14gam\uparrow Y\left\{ \begin{aligned}
& C{{O}_{2}}(a mol);NO(b mol) \\
& {{N}_{2}}(c mol);{{H}_{2}} \\
\end{aligned} \right. \\
& {{H}_{2}}O \\
\end{aligned} \right.$
$\xrightarrow{BT.S}(0,5y+0,6425)=0,715\to y=0,145\to {{m}_{A}}=82,395$
${{n}_{OH(\downarrow )}}={{n}_{NaOH}}-{{n}_{N{{H}_{3}}}}=1,285-0,025=1,26mol$
$\to {{m}_{kim loai (\downarrow )}}={{m}_{\downarrow }}-{{m}_{OH(\downarrow )}}=46,54-17.1,26=25,12gam$
${{n}_{NH_{4}^{+}}}={{n}_{N{{H}_{3}}}}=0,025mol;{{n}_{N{{a}^{+}}(Z)}}={{n}_{NaN{{O}_{3}}}}=0,145$
$\to {{m}_{Z}}=25,12+23.0,145+18.0,025+96.0,715=97,545gam$
$\xrightarrow{BTKL}{{n}_{{{H}_{2}}O}}=\dfrac{31,36+(98.0,715+85.0,145)-97,545-5,14}{18}=0,615mol$
$\xrightarrow{BT.H}{{n}_{{{H}_{2}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}-2{{n}_{NH_{4}^{+}}}-{{n}_{{{H}_{2}}O}}=0,715-2.0,025-0,615=0,05mol$
$\to \left\{ \begin{aligned}
& {{m}_{Y}}=44a+30b+28c+2.0,05=5,14 \\
& \xrightarrow{BT.N}b+2c+0,025=0,145 \\
& {{n}_{Y}}=a+b+c+0,05=0,2 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,04 \\
& b=0,1 \\
& c=0,01 \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& \xrightarrow{BT.C}{{n}_{FeC{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,04 \\
& \xrightarrow{BT.O}{{n}_{F{{e}_{3}}{{O}_{4}}}}=\dfrac{2.0,04+0,1+0,615-3.0,145-3.0,04}{4}=0,06mol \\
\end{aligned} \right.$
$\to \%{{m}_{F{{e}_{3}}{{O}_{4}}(X)}}=\dfrac{232.0,06}{31,36}.100\%=44,39\%$
& Mg \\
& Fe \\
& F{{e}_{3}}{{O}_{4}} \\
& FeC{{O}_{3}} \\
\end{aligned} \right.}_{31,36gam}+A\left\{ \begin{aligned}
& {{H}_{2}}S{{O}_{4}}:x \\
& NaN{{O}_{3}}:y \\
\end{aligned} \right.\to \left| \begin{aligned}
& \\
& Z\left\{ \begin{aligned}
& N{{a}^{+}} \\
& M{{g}^{2+}} \\
& F{{e}^{2+}} \\
& F{{e}^{3+}} \\
& NH_{4}^{+} \\
& SO_{4}^{2-} \\
\end{aligned} \right.\left| \begin{aligned}
& \xrightarrow[1,285mol]{+NaOH}\left| \begin{aligned}
& 46,54gam\downarrow \left\{ \begin{aligned}
& Mg{{(OH)}_{2}} \\
& Fe{{(OH)}_{2}} \\
& Fe{{(OH)}_{3}} \\
\end{aligned} \right. \\
& N{{a}_{2}}S{{O}_{4}}:(0,5y+0,6425)(BT.Na) \\
& \uparrow N{{H}_{3}}:0,025mol \\
& {{H}_{2}}O \\
\end{aligned} \right. \\
& \xrightarrow{+BaC{{l}_{2}}}\downarrow BaS{{O}_{4}}:x=0,715 \\
\end{aligned} \right. \\
& 5,14gam\uparrow Y\left\{ \begin{aligned}
& C{{O}_{2}}(a mol);NO(b mol) \\
& {{N}_{2}}(c mol);{{H}_{2}} \\
\end{aligned} \right. \\
& {{H}_{2}}O \\
\end{aligned} \right.$
$\xrightarrow{BT.S}(0,5y+0,6425)=0,715\to y=0,145\to {{m}_{A}}=82,395$
${{n}_{OH(\downarrow )}}={{n}_{NaOH}}-{{n}_{N{{H}_{3}}}}=1,285-0,025=1,26mol$
$\to {{m}_{kim loai (\downarrow )}}={{m}_{\downarrow }}-{{m}_{OH(\downarrow )}}=46,54-17.1,26=25,12gam$
${{n}_{NH_{4}^{+}}}={{n}_{N{{H}_{3}}}}=0,025mol;{{n}_{N{{a}^{+}}(Z)}}={{n}_{NaN{{O}_{3}}}}=0,145$
$\to {{m}_{Z}}=25,12+23.0,145+18.0,025+96.0,715=97,545gam$
$\xrightarrow{BTKL}{{n}_{{{H}_{2}}O}}=\dfrac{31,36+(98.0,715+85.0,145)-97,545-5,14}{18}=0,615mol$
$\xrightarrow{BT.H}{{n}_{{{H}_{2}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}-2{{n}_{NH_{4}^{+}}}-{{n}_{{{H}_{2}}O}}=0,715-2.0,025-0,615=0,05mol$
$\to \left\{ \begin{aligned}
& {{m}_{Y}}=44a+30b+28c+2.0,05=5,14 \\
& \xrightarrow{BT.N}b+2c+0,025=0,145 \\
& {{n}_{Y}}=a+b+c+0,05=0,2 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,04 \\
& b=0,1 \\
& c=0,01 \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& \xrightarrow{BT.C}{{n}_{FeC{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,04 \\
& \xrightarrow{BT.O}{{n}_{F{{e}_{3}}{{O}_{4}}}}=\dfrac{2.0,04+0,1+0,615-3.0,145-3.0,04}{4}=0,06mol \\
\end{aligned} \right.$
$\to \%{{m}_{F{{e}_{3}}{{O}_{4}}(X)}}=\dfrac{232.0,06}{31,36}.100\%=44,39\%$
Đáp án C.