ta có : \(\dfrac{Fe}{m}\dfrac{+}{ }\dfrac{O}{15,04-m}\dfrac{\rightarrow}{ }\dfrac{hợpchất\left(x\right)gồm\left(Fe;O\right)}{15,04}\)
rồi đem hòa tan \(x\) bằng dung dịch \(H_2SO_4\)
\(\Rightarrow ptpư:\left(\overset{0}{Fe};\overset{0}{O}\right)\overset{ }{+}\overset{ }{H_2}\overset{+6}{S}\overset{ }{O_4}\overset{ }{\rightarrow}\overset{+3}{Fe_2}\left(\overset{+6}{S}\overset{-2}{O_4}\right)_3\overset{ }{+}\overset{_4}{S}\overset{-2}{O_2}\overset{ }{+}\overset{ }{H_2}\overset{-2}{O}\)
QT khử : \(\overset{0}{\dfrac{Fe}{x}}\overset{ }{\rightarrow}\overset{+3}{Fe}\overset{ }{+}\overset{ }{\dfrac{3e}{3x}}\)
QT OXH : \(\overset{+6}{\dfrac{S}{0,06}}\overset{ }{+}\overset{ }{\dfrac{2e}{0,12}}\overset{ }{\rightarrow}\overset{+4}{S}\) ; \(\overset{0}{\dfrac{O}{y}}\overset{ }{+}\overset{ }{\dfrac{2e}{2y}}\overset{ }{\rightarrow}\overset{-2}{O}\)
\(\Rightarrow\left\{{}\begin{matrix}56x+16y=15,04\\3y=0,12+2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,24\end{matrix}\right.\)
\(\Rightarrow m=0,2.56=11.2\)