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Câu hỏi: Cho 33,4 gam hỗn hợp rắn $\mathrm{X}$ gồm $\mathrm{Mg}, \mathrm{MgO}, \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}$ và $\mathrm{FeCO}_{3}$ vào dung dịch chứa $1,29~ \text{mol} \text{HCl}$ và $0,166 \text{HN}{{\text{O}}_{3}},$ khuấy đều cho các phản ứng xảy ra hoàn toàn, thu được dung dịch Y chỉ chứa các muối và 0,163 mol hỗn hợp khí $Z$ gồm $\mathrm{N}_{2} \mathrm{O}, \mathrm{N}_{2}$ và $0,1~ \text{mol} \text{ C}{{\text{O}}_{2}}$. Cho dung dịch $\text{AgN}{{\text{O}}_{\text{3}}}$ dư vào dung dịch Y thu được 191,595 gam kết tủa. Nếu tác dụng tối đa với các chất tan có trong dung dịch Y cần dùng dung dịch chứa $1,39~ \text{mol} \text{KOH}.$ Biết rằng tổng số mol nguyên tử oxi có trong $\mathrm{X}$ là 0,68 mol. Số $\mathrm{mol}$ của $\mathrm{N}_{2}$ có trong $\mathrm{Z}$ là
A. $0,028$.
B. $0,031$.$$
C. $0,033$.
D. $0,035$.
A. $0,028$.
B. $0,031$.$$
C. $0,033$.
D. $0,035$.
$39,4 X\left\{ \begin{aligned}
& \text{Mg:} \text{x} \\
& \text{MgO:} \text{y} \\
& \text{Fe}{{\left( \text{N}{{\text{O}}_{\text{3}}} \right)}_{\text{2}}}\text{:} \text{z} \\
& \text{FeC}{{\text{O}}_{\text{3}}}\text{:0,1} \left( {{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}} \right) \\
\end{aligned} \right.\xrightarrow[\text{HN}{{\text{O}}_{\text{3}}}: 0,166]{\text{HCl:} 1,29}\left\{ \begin{aligned}
& 0,163 \text{Z}\left\{ \begin{aligned}
& {{\text{N}}_{\text{2}}}\text{O} \\
& {{\text{N}}_{\text{2}}} \\
& \text{C}{{\text{O}}_{\text{2}}}\text{:0,1} \\
\end{aligned} \right. \\
& Y\xrightarrow{\text{AgN}{{\text{O}}_{3}}}191,595\left\{ \begin{aligned}
& \text{Ag} \\
& \text{AgCl:} \text{1,29} \left( {{\text{n}}_{\text{HCl}}} \right) \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\to {{n}_{Ag}}=\dfrac{191,595-1,29.\left( 108+35,5 \right)}{108}=0,06\to {{n}_{F{{e}^{2+}}\left( Y \right)}}=0,06$
$Y\left\{ \begin{aligned}
& \text{F}{{\text{e}}^{\text{2+}}}\text{:} \text{0,06} \\
& \text{F}{{\text{e}}^{\text{3+}}}\text{:} \text{BT} \text{Fe:} {{\text{n}}_{\text{F}{{\text{e}}^{\text{3+}}}}}\text{=}\left( \text{z+0,1-0,06} \right)\text{=z+0,04} \\
& \text{M}{{\text{g}}^{\text{2+}}}\text{:} \text{x+y} \\
& \text{C}{{\text{l}}^{\text{-}}}\text{:1} \text{,29} \\
& \text{NO}_{\text{3}}^{\text{-}}\text{:} {{\text{n}}_{\text{NO}_{\text{3}}^{\text{-}}}}\text{+}{{\text{n}}_{\text{C}{{\text{l}}^{\text{-}}}}}\text{=}{{\text{n}}_{\text{ion+}\left( \text{Y} \right)}}\text{=}{{\text{n}}_{\text{KOH}}}\text{=1,39}\to {{\text{n}}_{\text{NO}_{\text{3}}^{\text{-}}}}=0,1 \\
& \text{NH}_{\text{4}}^{\text{+}}\text{:} \text{BT} \text{N:}{{\text{n}}_{\text{NH}_{\text{4}}^{\text{+}}}}=\text{2}{{\text{n}}_{\text{Fe}{{\left( \text{N}{{\text{O}}_{\text{3}}} \right)}_{\text{2}}}}}\text{+}{{\text{n}}_{\text{HN}{{\text{O}}_{\text{3}}}}}\text{-2}{{\text{n}}_{{{\text{N}}_{\text{2}}}\text{O,}{{\text{N}}_{\text{2}}}}}\text{-}{{\text{n}}_{\text{NO}_{\text{3}}^{\text{-}}}}=\left( 2\text{z}+0,166-0,063.2-0,1 \right)=2\text{z}-0,06 \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& y+6\text{z}+0,1.3=0,68 \left( {{n}_{O\left( X \right)}} \right) \\
& 0,06.2+3.\left( \text{z}+0,04 \right)+2.\left( x+y \right)+\left( 2\text{z}-0,06 \right)=1,39\left( {{n}_{KOH}} \right) \\
& 24x+40y+180\text{z}+11,6=33,4 \\
\end{aligned} \right. \to \left\{ \begin{aligned}
& x=0,4 \\
& y=0,08 \\
& \text{z}=0,05 \\
\end{aligned} \right.$
BT H: ${{n}_{{{H}_{2}}O}}=0,648$
BT O: ${{n}_{{{N}_{2}}O}}=0,03$
$\to {{n}_{{{N}_{2}}}}=0,033$
& \text{Mg:} \text{x} \\
& \text{MgO:} \text{y} \\
& \text{Fe}{{\left( \text{N}{{\text{O}}_{\text{3}}} \right)}_{\text{2}}}\text{:} \text{z} \\
& \text{FeC}{{\text{O}}_{\text{3}}}\text{:0,1} \left( {{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}} \right) \\
\end{aligned} \right.\xrightarrow[\text{HN}{{\text{O}}_{\text{3}}}: 0,166]{\text{HCl:} 1,29}\left\{ \begin{aligned}
& 0,163 \text{Z}\left\{ \begin{aligned}
& {{\text{N}}_{\text{2}}}\text{O} \\
& {{\text{N}}_{\text{2}}} \\
& \text{C}{{\text{O}}_{\text{2}}}\text{:0,1} \\
\end{aligned} \right. \\
& Y\xrightarrow{\text{AgN}{{\text{O}}_{3}}}191,595\left\{ \begin{aligned}
& \text{Ag} \\
& \text{AgCl:} \text{1,29} \left( {{\text{n}}_{\text{HCl}}} \right) \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\to {{n}_{Ag}}=\dfrac{191,595-1,29.\left( 108+35,5 \right)}{108}=0,06\to {{n}_{F{{e}^{2+}}\left( Y \right)}}=0,06$
$Y\left\{ \begin{aligned}
& \text{F}{{\text{e}}^{\text{2+}}}\text{:} \text{0,06} \\
& \text{F}{{\text{e}}^{\text{3+}}}\text{:} \text{BT} \text{Fe:} {{\text{n}}_{\text{F}{{\text{e}}^{\text{3+}}}}}\text{=}\left( \text{z+0,1-0,06} \right)\text{=z+0,04} \\
& \text{M}{{\text{g}}^{\text{2+}}}\text{:} \text{x+y} \\
& \text{C}{{\text{l}}^{\text{-}}}\text{:1} \text{,29} \\
& \text{NO}_{\text{3}}^{\text{-}}\text{:} {{\text{n}}_{\text{NO}_{\text{3}}^{\text{-}}}}\text{+}{{\text{n}}_{\text{C}{{\text{l}}^{\text{-}}}}}\text{=}{{\text{n}}_{\text{ion+}\left( \text{Y} \right)}}\text{=}{{\text{n}}_{\text{KOH}}}\text{=1,39}\to {{\text{n}}_{\text{NO}_{\text{3}}^{\text{-}}}}=0,1 \\
& \text{NH}_{\text{4}}^{\text{+}}\text{:} \text{BT} \text{N:}{{\text{n}}_{\text{NH}_{\text{4}}^{\text{+}}}}=\text{2}{{\text{n}}_{\text{Fe}{{\left( \text{N}{{\text{O}}_{\text{3}}} \right)}_{\text{2}}}}}\text{+}{{\text{n}}_{\text{HN}{{\text{O}}_{\text{3}}}}}\text{-2}{{\text{n}}_{{{\text{N}}_{\text{2}}}\text{O,}{{\text{N}}_{\text{2}}}}}\text{-}{{\text{n}}_{\text{NO}_{\text{3}}^{\text{-}}}}=\left( 2\text{z}+0,166-0,063.2-0,1 \right)=2\text{z}-0,06 \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& y+6\text{z}+0,1.3=0,68 \left( {{n}_{O\left( X \right)}} \right) \\
& 0,06.2+3.\left( \text{z}+0,04 \right)+2.\left( x+y \right)+\left( 2\text{z}-0,06 \right)=1,39\left( {{n}_{KOH}} \right) \\
& 24x+40y+180\text{z}+11,6=33,4 \\
\end{aligned} \right. \to \left\{ \begin{aligned}
& x=0,4 \\
& y=0,08 \\
& \text{z}=0,05 \\
\end{aligned} \right.$
BT H: ${{n}_{{{H}_{2}}O}}=0,648$
BT O: ${{n}_{{{N}_{2}}O}}=0,03$
$\to {{n}_{{{N}_{2}}}}=0,033$
Đáp án C.