Đáp án đúng: B
30,97%.
Phân tích : Đặt $\displaystyle {{\text{n}}_{\text{Cu}}}\text{=}{{\text{n}}_{\text{CuO}}}\text{=a; }{{\text{n}}_{\text{Cu(N}{{\text{O}}_{\text{3}}}\text{)}}}\text{=b}\text{.}$Khi cho X tan hoàn toàn trong dung dịch H2SO4 thì ta chỉ thu được dung dịch chỉ chứa một chất tan duy nhất nên sau phản ứng thì$\displaystyle \text{NO}_{\text{3}}^{\text{-}}$ hết và muối thu được là CuSO4 .
Ta có :$\displaystyle \text{CuO+2}{{\text{H}}^{\text{+}}}\xrightarrow{{}}\text{C}{{\text{u}}^{\text{2+}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}$
$\begin{array}{l}\text{3Cu}\,\text{+}\,\text{8}{{\text{H}}^{\text{+}}}\,\text{+}\,\text{2NO}_{\text{3}}^{\text{-}}\,\xrightarrow{{}}\,\text{C}{{\text{u}}^{\text{2+}}}\,\text{+}\,\text{2NO}\,\text{+}\,\text{4}{{\text{H}}_{\text{2}}}\text{O}\\\,\,\text{a}\,\,\,\,\,\,\,\,\,\,\frac{\text{8a}}{\text{3}}\,\,\,\,\,\,\,\,\,\,\,\text{2b}\end{array}$
$\Rightarrow \,\text{2a}\,\text{+}\,\frac{\text{8a}}{\text{3}}\,\text{=}\,{{\text{n}}_{{{\text{H}}^{\text{+}}}}}\,\text{=}\,\text{2}{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\,\text{=}\,\text{1}\text{,4}\,\Rightarrow \,\text{a}\,\text{=}\,\text{0}\text{,3}$
Mà$\displaystyle \text{2b=}\frac{\text{2}}{\text{3}}\text{a}\Rightarrow \text{b=0,1}$Vậy khối lượng Cu trong X là :$\displaystyle \frac{\text{0,3}\text{.64}}{\text{0,3}\text{.(64+80)+0,1}\text{.188}}\text{=30,97}{\scriptstyle{}^{\text{o}}\!\!\diagup\!\!{}_{\text{o}}\;}$
