$2A{{l}_{2}}{{O}_{3}}+3C\to 4Al+3C{{O}_{2}}$  $?kgC={{10}^{3}}kgAl\times \frac{{{10}^{3}}gAl}{1kgAl}\times \frac{1molAl}{27gAl}\times \frac{3molC}{4molAl}\times \frac{12gC}{1molC}\times \frac{1kgC}{1000gC}=333$  $?{{m}^{3}}C{{O}_{2}}={{10}^{3}}kgAl\times \frac{{{10}^{3}}gAl}{1kgAl}\times \frac{1molAl}{27gAl}\times \frac{3molC{{O}_{2}}}{4molAl}\times \frac{25LC{{O}_{2}}}{1molC{{O}_{2}}}\times \frac{1{{m}^{3}}}{1000L}=694/4$