`n_X = (12,395.10^3)/(24,79) = 500 (mol)`

`m_{giảm} = m_C = 4,8 (kg)=4800(g)`

`M_X = 16.2 = 32 (g//mol)`

`m_X = 32.500 = 16000(g)`

`=> m_{O_2} = m_X - m_C = 16000-4800 = 11200 (g)`

`=> n_{O_2} = (11200)/(32) = 350 (mol)`

Bảo toàn e: `n_{Al} = 4/3 n_{O_2} = (1400)/3 (mol)`

`=> m = m_{Al} = (1400)/3 . 27 = 12600(g) = 12,6 (kg)`