\(\dfrac{sốphântửCaO}{sốphântửAl_2O_3}=\dfrac{1}{2}\Rightarrow\dfrac{m_{CaO}}{m_{Al_2O_3}}=\dfrac{1\times56}{2\times102}=\dfrac{14}{51}\)
\(\Rightarrow m_{CaO}=8\div\left(14+51\right)\times14=1,72\left(tấn\right)\)
\(\Rightarrow m_{Al_2O_3}=8-1,72=6,28\left(tấn\right)\)