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Câu hỏi: Cho dung dịch Ba(OH)2 dư vào 50 ml dung dịch X chứa các ion: $NH_{4}^{+},SO_{4}^{2-},NO_{3}^{-}$ đun nóng, thu được 11,65 gam kết tủa và 4,48 lít khí Y thoát ra (đktc). Nồng độ mol mỗi muối trong dung dịch X là:
A. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 1M;N{{H}_{4}}N{{O}_{3}} 2M.$
B. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 1M;N{{H}_{4}}N{{O}_{3}} 1M.$
C. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 2M;N{{H}_{4}}N{{O}_{3}} 2M.$
D. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 1M;N{{H}_{4}}N{{O}_{3}} 0,5M.$
A. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 1M;N{{H}_{4}}N{{O}_{3}} 2M.$
B. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 1M;N{{H}_{4}}N{{O}_{3}} 1M.$
C. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 2M;N{{H}_{4}}N{{O}_{3}} 2M.$
D. ${{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} 1M;N{{H}_{4}}N{{O}_{3}} 0,5M.$
$\underbrace{X\left\{ \begin{aligned}
& {{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} \\
& N{{H}_{4}}N{{O}_{3}} \\
\end{aligned} \right.}_{0,05\left( l \right)}\leftrightarrow X\left\{ \begin{aligned}
& NH_{4}^{+} \\
& SO_{4}^{2-} \\
& NO_{3}^{-} \\
\end{aligned} \right.\xrightarrow{+Ba{{\left( OH \right)}_{2}}}\left\{ \begin{aligned}
& BaS{{O}_{4}}\downarrow :0,05\left( mol \right) \\
& N{{H}_{3}}\uparrow :0,2\left( mol \right) \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& {{n}_{{{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}}}}={{n}_{SO_{4}^{2-}}}=0,05\left( mol \right) \\
& {{n}_{N{{H}_{4}}N{{O}_{3}}}}=\underbrace{{{n}_{N{{H}_{3}}}}}_{0,2}-\underbrace{2{{n}_{{{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}}}}}_{2.0,05}=0,1\left( mol \right) \\
\end{aligned} \right.\leftrightarrow \left\{ \begin{aligned}
& \left[ {{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} \right]=1M \\
& \left[ N{{H}_{4}}N{{O}_{3}} \right]=2M \\
\end{aligned} \right.$
& {{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} \\
& N{{H}_{4}}N{{O}_{3}} \\
\end{aligned} \right.}_{0,05\left( l \right)}\leftrightarrow X\left\{ \begin{aligned}
& NH_{4}^{+} \\
& SO_{4}^{2-} \\
& NO_{3}^{-} \\
\end{aligned} \right.\xrightarrow{+Ba{{\left( OH \right)}_{2}}}\left\{ \begin{aligned}
& BaS{{O}_{4}}\downarrow :0,05\left( mol \right) \\
& N{{H}_{3}}\uparrow :0,2\left( mol \right) \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& {{n}_{{{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}}}}={{n}_{SO_{4}^{2-}}}=0,05\left( mol \right) \\
& {{n}_{N{{H}_{4}}N{{O}_{3}}}}=\underbrace{{{n}_{N{{H}_{3}}}}}_{0,2}-\underbrace{2{{n}_{{{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}}}}}_{2.0,05}=0,1\left( mol \right) \\
\end{aligned} \right.\leftrightarrow \left\{ \begin{aligned}
& \left[ {{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}} \right]=1M \\
& \left[ N{{H}_{4}}N{{O}_{3}} \right]=2M \\
\end{aligned} \right.$
Đáp án A.