Câu 86:

$n_{KMnO_4}=\dfrac{47,4}{158}=0,3(mol)\\ n_{SO_2}=\dfrac{3,92}{22,4}=0,175(mol)\\ PTHH:2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \text{Bảo toàn KL: }m_{O_2}=m_{KMnO_4}-m_{\text{hh rắn}}=47,4-44,2=3,2(g)\\ \Rightarrow n_{O_2}=\dfrac{3,2}{32}=0,1(mol)\\ \text{Bảo toàn KL: }m_{KL}=m_{oxit}-m_{O_2}=21,2-3,2=18(g)\\ \text{Đặt }(n_{Cu};n_{Fe})=(x;y)(mol)\\ \Rightarrow 64x+56y=18(1)\\ Cu\to 2e+Cu^{2+}\\ x\longrightarrow 2x(mol)\\ Fe\to 3e+Fe^{3+}\\ y\longrightarrow 3y(mol)\\ O_2+4e\to 2O^{2-}\\ 0,1\rightarrow 0,4(mol)\\ S^{6+}+2e\to S^{+4}\\ 0,35\leftarrow 0,175(mol)\\ \text{Bảo toàn e: }\begin{cases} 64x+56y=18\\ 2x+3y=0,75 \end{cases}\Rightarrow x=y=0,15(mol)\\ \Rightarrow \begin{cases} \%m_{Fe}=\dfrac{0,15.56}{18}.100\%=46,67\%\\ \%m_{Cu}=100\%-46,67\%=53,33\% \end{cases}$