Đáp án đúng: A
43,44%
Quy đổi
$X\left\{ \begin{array}{l}KMn{{O}_{4}}\,\,a\\KCl\,\,\,\,b\\CaC{{l}_{2}}\,\,\,c\\O\,\,\,(1,2b\,+\,2,4c)\end{array} \right.\,\xrightarrow{{{t}^{o}}}\,\,\left\{ \begin{array}{l}{{O}_{2}}\,\,(m\,\,gam)\\Y\,\,\,(m\,+\,23,34\,\,gam)\end{array} \right.$
⇒ mX = 2m + 23,34
${{n}_{HCl}}\,=\,0,76;\,\,{{n}_{C{{l}_{2}}}}\,=\,0,26\,\Rightarrow \,{{n}_{O\,\,(trong\,Y)}}\,=\,{{n}_{{{H}_{2}}O}}=0,38;\,{{n}_{AgCl}}\,=\,0,44$
$40,32\,gam\,T\left\{ \begin{array}{l}KCl\,\,(a+b)\\CaC{{l}_{2}}\,\,c\\MnC{{l}_{2}}\,\,a\end{array} \right.\,\begin{matrix} \Rightarrow m={{m}_{{{O}_{2}}}}=16.(4a+1,2b+2,4c-0,38) \\ =64a+19,2b+38,4c-6,08\,\,\, \\ \end{matrix}$
Ta có
$\left\{ \begin{array}{l}a+b+2c+2a=0,44\\b+2c+0,76-0,26.2=0,44\\158a+93,7b+149,4c=2(64a+19,2b+38,4c-6,08)\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}a=0,08\\b=0,08\\c=0,06\end{array} \right.$
$\Rightarrow {\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}KMn{{O}_{4}}=\,\frac{0,08.158}{29,1}=43,436{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}$