- Chọn B.

- Khối lượng $Fe{{S}_{2}}$có trong quặng là: ${{m}_{Fe{{S}_{2}}}}=100-\frac{4}{100}\times 100=96\left( gam \right)$

- Ta có: ${{n}_{Fe{{S}_{2}}}}=\frac{96}{120}=0,8\left( mol \right)$.

4$Fe{{S}_{2}}$      +           11${{O}_{2}}$   $\to $ 2$F{{e}_{2}}{{O}_{3}}$        +   8$S{{O}_{2}}$

0,8(mol)        $\to $                                            x(mol)

$x=0,8\times \frac{8}{4}=2\left( mol \right)$.${{V}_{S{{O}_{2}}}}=2\times 22,4=44,8\left( lit \right)$