Cho m gam hỗn hợp X gồm Cu và Fe2O3 vào 200,0 ml dung dịch chứa NaNO3

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} +\left\{\begin{array}{l}\mathrm{Dung} \mathrm{dịch} \mathrm{Y} \mathrm{có} {\mathrm{Fe}}^{2+} \mathrm{nên} \mathrm{cho} \mathrm{HCl} \mathrm{vào} \mathrm{có} \mathrm{NO} \mathrm{bay} \mathrm{ra}\\ \sum {\mathrm{n}}_{\mathrm{NO}}=0,1<{\mathrm{n}}_{{\mathrm{NaNO}}_3}=0,2 \mathrm{nên} \mathrm{cuối} \mathrm{cùng} \mathrm{dung} \mathrm{dịch} \mathrm{còn} {{\mathrm{NO}}_3}^{-}\end{array}\right. \\ + \mathrm{Bản} \mathrm{chất} \mathrm{phản} \mathrm{ứng}: \\ \underset{\mathrm{X}}{\underbrace{\left\{\mathrm{Cu}: \mathrm{x} \\ {\mathrm{Fe}}_2{\mathrm{O}}_3: \mathrm{yl}\right\}}}\underset{{\mathrm{NaNO}}_3: 0,2}{\overset{{\mathrm{H}}_2{\mathrm{SO}}_4: 0,4}{\to }}\underset{0,05 \mathrm{mol}}{\underbrace{\mathrm{NO}\uparrow }}+\underset{\mathrm{Y}}{\underbrace{\left\{{\mathrm{Cu}}^{2+}{\mathrm{Fe}}^{2+}, {\mathrm{Fe}}^{3+}\mathrm{l} \\ {{\mathrm{NO}}_3}^{-}, {{\mathrm{SO}}_4}^{2-}\right\}}}+{\mathrm{H}}_2\mathrm{O} \\ + \mathrm{BTNT} \mathrm{H}: 0,4.2=0,05.4+2.3\mathrm{y}\Rightarrow \mathrm{y}=0,1. \\ \underset{\mathrm{Y}}{\underbrace{\left\{{\mathrm{Cu}}^{2+}{\mathrm{Fe}}^{2+}, {\mathrm{Fe}}^{3+} \\ {{\mathrm{NO}}_3}^{-}, {{\mathrm{SO}}_4}^{2-}\right\}}}\stackrel{\mathrm{HCl}}{\to }\underset{0,05 \mathrm{mol}}{\underbrace{\mathrm{NO}\uparrow }}+ \left\{{\mathrm{Fe}}^{3+}, {\mathrm{Cu}}^{2+} \\ {\mathrm{Na}}^{+}, {{\mathrm{SO}}_4}^{2-} \\ {{\mathrm{NO}}_3}^{-}\right\}+{\mathrm{H}}_2\mathrm{O} \\ + \mathrm{BTE}: 2{\mathrm{n}}_{\mathrm{Cu}}=3{\mathrm{n}}_{\mathrm{NO}}\Rightarrow {\mathrm{n}}_{\mathrm{Cu}}=\frac{3(0,05+0,05)}{2}=0,15. \\ + \mathrm{m}=0,1.160+0,15.64=\boxed{25,6 \mathrm{gam}} \end{gathered}$$