Đáp án đúng: C
8,69%.
Dung dịch C có phản ứng với Zn cho khí H2, vậy trong C có axit ⟹ B phải là H2SO4 hoặc muối M(HSO4)n
Mà chất ban đầu phản ứng với nước nên có thể là SO3, hoặc H2SO4.nSO3
$\displaystyle \text{BaC}{{\text{l}}_{\text{2}}}\text{+ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4 }\!\!~\!\!\text{ }}}\xrightarrow{{}}\text{BaS}{{\text{O}}_{\text{4}}}\text{+2HCl }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\left( \text{1} \right)$
$\displaystyle \text{Zn + 2HCl }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ZnC}{{\text{l}}_{\text{2 }\!\!~\!\!\text{ }}}\text{+ }{{\text{H}}_{\text{2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}\left( \text{2} \right)$
$\displaystyle \text{Zn + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\xrightarrow{{}}\text{ZnS}{{\text{O}}_{\text{4}}}\text{+ }{{\text{H}}_{\text{2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}\left( \text{3} \right)$
Ta có$\displaystyle {{\text{n}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\text{ }\!\!~\!\!\text{ = }\!\!~\!\!\text{ }{{\text{n}}_{{{\text{H}}_{\text{2}}}}}\text{=}\frac{\text{1}\text{,792}}{\text{22}\text{,4}}\text{=0}\text{,08}\,\,\text{mol}$
* TH1: A là SO3 $\displaystyle \Rightarrow {{\text{n}}_{\text{SO}}}_{_{\text{3}}}\text{=}\frac{\text{6}\text{,58}}{\text{80}}\text{=0}\text{,08225}
e \text{0}\text{,08}$(Loại)
* TH2: A là H2SO4.nSO3
$\displaystyle {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{.nS}{{\text{O}}_{\text{3}}}\text{+n}{{\text{H}}_{\text{2}}}\text{O}\xrightarrow{{}}\left( \text{n+1} \right)\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$
Ta có$\displaystyle \frac{\text{6}\text{,58}}{\text{98}+\text{80n}}=\frac{\text{0}\text{,08}}{\text{n}+\text{1}}\Rightarrow ~~n~=\text{ }7$
⟹ A là H2SO4.7H2O
Khối lượng dung dịch D là
$\displaystyle {{\text{m}}_{\text{BaC}{{\text{l}}_{\text{2}}}}}\text{=}\frac{\text{4}\text{,66}\text{.208}}{\text{233}}\text{=}\,\text{4}\text{,16gam}$
mdd = 6,58 +100 + 4,16 + 0,08.65 – 0,08.2 – 4,66 = 111,12 gam
$\displaystyle {{\text{n}}_{\text{ZnC}{{\text{l}}_{\text{2}}}}}\text{= }{{\text{n}}_{\text{BaS}{{\text{O}}_{\text{4}}}}}\text{= 0}\text{,2 mol}$
$\displaystyle {{\text{n}}_{\text{ZnS}{{\text{O}}_{\text{4}}}}}\text{= 0}\text{,08 - 0}\text{,02 = 0}\text{,06 mol}$
$\displaystyle \text{C}{{{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}}_{\text{ZnC}{{\text{l}}_{\text{2}}}}}\text{=}\frac{\text{0}\text{,02}\text{.136}}{\text{111}\text{,12}}\text{.100}\,\text{=}\,\text{2}\text{,45}{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}$$\displaystyle \text{C}{{{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}}_{\text{ZnS}{{\text{O}}_{\text{4}}}}}\text{=}\frac{\text{0}\text{,06}\text{.161}}{\text{111}\text{,12}}\text{.100}\,\text{=}\,\text{8}\text{,69}{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}$
