22/
$m_{Al}= 1T= 10^6 (g)$
$2Al_2O_3\to 4Al+3O_2$
Cứ 204g oxit tạo 108g nhôm.
$\Rightarrow m_{Al_2O_3}=\frac{10^6.204}{108}=\text{ 1 888 888,89} (g)$
$\Rightarrow m_{\text{boxit}}=\text{1 888 888,89}:50\% =\text{ 3 777 777,78}$
$m= \frac{AIt}{nF}$
$\Leftrightarrow 10^6=\frac{27.30000.t}{3.96500}$
$\Leftrightarrow t=357407s$