- Chọn D.

- Ta có: ${{n}_{ZnS}}=\frac{9,7}{97}=0,1\left( mol \right);{{n}_{{{O}_{2}}}}=\frac{4,48}{22,4}=0,2\left( mol \right)$.

ZnS      +           ${{O}_{2}}$   $\to $ ZnO        +   $S{{O}_{2}}$

0,1(mol)        0,2(mol)                        x(mol)

Do ${{n}_{ZnS}}<{{n}_{{{O}_{2}}}}\to $${{O}_{2}}$còn dư sau phản ứng.

${{n}_{S{{O}_{2}}}}=x={{n}_{ZnS}}=0,1\left( mol \right)$. ${{V}_{S{{O}_{2}}}}=0,1\times 22,4=2,24\left( lit \right)$