$\rm n_{Al_2O_3}=\dfrac{1.60\%}{102}=\dfrac{1}{170}(tấn~mol)$ 

Ta có $\rm n_{Al}=2n_{Al_2O_3}=\dfrac{2}{170}(tấn~mol)$ 

`->` $\rm m_{Al}=\dfrac{2}{170}.27≈0,318(tấn)$