\(2Al_2O_3\left(\dfrac{m.10^6}{170}\right)\underrightarrow{ĐPNC}4Al\left(\dfrac{m.10^6}{85}\right)+3O_2\)
\(m_{Al_2O_3}=0,6m.10^6\left(g\right)\)
\(\Rightarrow n_{Al_2O_3}=\dfrac{0,6m.10^6}{102}=\dfrac{m.10^6}{170}\left(mol\right)\)
\(\Rightarrow\dfrac{m.10^6}{85}.27=54.10^6\)
\(\Leftrightarrow m=170\left(tan\right)\)