Hướng dẫn:

$CuO+{{H}_{2}}\xrightarrow{{{t}^{0}}}Cu+{{H}_{2}}O$

$FeO+{{H}_{2}}\xrightarrow{{{t}^{0}}}Fe+{{H}_{2}}O$

$F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\xrightarrow{{{t}^{0}}}2Fe+3{{H}_{2}}O$

Theo định luật bảo toàn khối lượng ta có: ${{m}_{A}}+{{m}_{{{H}_{2}}}}={{m}_{KL}}+{{m}_{{{H}_{2}}O}}$

$\Rightarrow {{m}_{{{H}_{2}}}}={{m}_{KL}}+{{m}_{{{H}_{2}}O}}-{{m}_{A}}=16,05-15,75$

$\Rightarrow {{m}_{{{H}_{2}}}}=0,3\left( g \right)$

$\Rightarrow {{n}_{{{H}_{2}}}}=\frac{{{m}_{{{H}_{2}}}}}{{{M}_{{{H}_{2}}}}}=\frac{0,3}{2}=0,15\left( mol \right)$

$\Rightarrow {{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,15=3,36\left( l \right)$

$\Rightarrow $ Chọn đáp án A