\(C_6H_{12}O_6\rightarrow2C_2H_5OH+2CO_2\\ n_{C_6H_{12}O_6}=\dfrac{16,2}{180}=0,09\left(mol\right)\\ n_{C_2H_5OH}=2n_{C_6H_{12}O_6}=0,18\left(mol\right)\\ \Rightarrow m_{C_2H_5OH}=0,18.46=8,28\left(kg\right)\\ VìH=80\%\Rightarrow m_{C_2H_5OH}=8,28.80\%=6,624\left(g\right)\)