\(2xAl+yO_2\rightarrow2Al_xO_y\)
0,1__________0,1/x
Ta có :
\(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)
\(M_{AlxOy}=\frac{5,1}{\frac{0,1}{x}}\)
\(\Leftrightarrow M_{AlxOy}=51x\)
\(27x+16y=51\Leftrightarrow16y=24x\)
\(\Leftrightarrow\frac{x}{y}=\frac{2}{3}\)
Vậy CTHH là Al2O3