Cho 7,65 gam hỗn hợp X gồm Al và Al2O3 tan hoàn toàn trong dung dịch Y

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} + \mathrm{Ta} \mathrm{có}: {\mathrm{n}}_{\mathrm{Al}}=0,17 \mathrm{mol}; {\mathrm{n}}_{{\mathrm{Al}}_2{\mathrm{O}}_3}=0,03 \mathrm{mol}; {\mathrm{n}}_{{\mathrm{BaSO}}_4}=0,4 \mathrm{mol}; {\mathrm{n}}_{\mathrm{NaOH}}=0,935 \mathrm{mol}. \\ + \mathrm{Sơ} \mathrm{đồ} \mathrm{phản} \mathrm{ứng}: \\ \underset{\mathrm{X}}{\underbrace{\left\{\mathrm{Al} \\ {\mathrm{Al}}_2{\mathrm{O}}_3\right\}}}+ \underset{\mathrm{Y}}{\underbrace{\left\{{\mathrm{H}}_2{\mathrm{SO}}_4 \\ {\mathrm{NaNO}}_3\right\}}}\stackrel{}{\to }\underset{\mathrm{Z}}{\underbrace{\left\{{{\mathrm{NH}}_4}^{+}, {\mathrm{Al}}^{3+} \\ {\mathrm{Na}}^{+}, {{\mathrm{SO}}_4}^{2-}\right\}}}+ \underset{\mathrm{T}}{\underbrace{\left\{{\mathrm{H}}_2\uparrow \\ {\mathrm{NO}}_{\overline{\mathrm{x}}}\uparrow \right\}}} \\ \downarrow \\ \left\{\begin{array}{l}\stackrel{{\mathrm{BaCl}}_2 \mathrm{dư}}{\to }{\mathrm{BaSO}}_4\\ \stackrel{\mathrm{NaOH} \mathrm{pư} \max }{\to }\underset{\mathrm{W}}{\underbrace{\left\{{{\mathrm{AlO}}_2}^{-} \\ {{\mathrm{SO}}_4}^{2-}, {\mathrm{Na}}^{+}\right\}}}\end{array}\right. \\ + \mathrm{Trong} \mathrm{W} \left\{\mathrm{BT} \mathrm{Al}: {\mathrm{n}}_{{{\mathrm{AlO}}_2}^{-}}=0,23 \\ \mathrm{BT} \mathrm{S}: {\mathrm{n}}_{{{\mathrm{SO}}_4}^{2-}}=0,4\right.\Rightarrow \left\{\mathrm{BTĐT} \mathrm{trong} \mathrm{W}: {\mathrm{n}}_{{\mathrm{Na}}^{+}}=1,03 \\ \mathrm{BT} \mathrm{Na}: {\mathrm{n}}_{{\mathrm{Na}}^{+} \mathrm{trong} \mathrm{Y}}={\mathrm{n}}_{{\mathrm{Na}}^{+} \mathrm{trong} \mathrm{Z}}=0,095\right. \\ + \left\{\mathrm{BTĐT} \mathrm{trong} \mathrm{Z}: {\mathrm{n}}_{{{\mathrm{NH}}_4}^{+}}=0,015; \mathrm{BT} \mathrm{N}: {\mathrm{n}}_{{\mathrm{NO}}_{\mathrm{x}}}=0,08 \\ \mathrm{BT} \mathrm{E}: 3.0,17=0,08(5-2\overline{\mathrm{x}})+0,015.2+0,015.8\right.\Rightarrow \left\{\overline{\mathrm{x}}=0,25 \\ {\mathrm{m}}_{\mathrm{T}}=1,47\approx \boxed{1,5 \mathrm{gam}}\right. \end{gathered}$$