$\rm n_{Al}=\dfrac{27}{27}=1$ (tấn mol)

$\rm 2Al_2O_3\ \rightarrow\ 4Al +3O_2$

`=>` $\rm n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}$ (tấn mol)

$\bullet\ \ \rm C\ +\ O_2\ \rightarrow\ CO_2$

$\bullet\ \ \rm 2C\ +\ O_2\ \rightarrow\ 2CO$

Gọi số mol thu được ở anode là `x` (tấn mol)

`=> \text{n}_{\text{CO}_2}=0,8 \x ; \text{n}_{\text{CO}}=0,1x; \text{n}_{\text{O}_2} (\text{dư}) =0,1x`

`=> \text{n}_{\text{O}_2}=0,8x+0,05x+0,1x=3/4 => x=15/19`

`=> \text{n}_{\text{C}}=0,8x+0,1x=0,9. 15/19=27/38` (tấn mol)

`=> \text{m}_{\text{C}}=27/38 .12=8,53 ~~9` (tấn)