$CuO+{{H}_{2}}S{{O}_{4}}\to CuS{{O}_{4}}+{{H}_{2}}O$  

Ban đầu:              0,04     0,02                                        (mol)

Phản ứng:             0,02  $\leftarrow $0,02                      (mol)

Sau phản ứng:      0,02        0                                          (mol)

${{n}_{CuO}}=\frac{{{m}_{CuO}}}{{{M}_{CuO}}}=\frac{3,2}{80}=0,04$ (mol)

${{n}_{{{H}_{2}}S{{O}_{4}}}}={{V}_{{{H}_{2}}S{{O}_{4}}}}.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,1.0,2=0,02$ (mol)

Theo phương trình, CuO và ${{H}_{2}}S{{O}_{4}}$ phản ứng theo tỉ lệ 1:1

${{n}_{CuO}}>{{n}_{{{H}_{2}}SO}}_{_{4}}\Rightarrow CuO$ dư

_{$\Rightarrow $${{n}_{Cu{{O}_{du}}}}={{n}_{CuO}}-{{n}_{{{H}_{2}}S{{O}_{4}}}}=0,04-0,02=0,02\left( mol \right)$}

_{$\Rightarrow $${{m}_{Cu{{O}_{d\text{u}}}}}={{n}_{Cu{{O}_{du}}}}.{{M}_{CuO}}=0,02.80=1,6\left( g \right)$}

$\Rightarrow $Chọn đáp án B