Hòa tan hoàn toàn 1,62 gam Al trong 280 ml dung dị...

B

- Hướng dẫn giải

\(+ \,\,BTE:\,\,{n_{NO}} = {n_{Al}} = 0,06\)

\( \Rightarrow \underbrace {Al}_{0,06\,\,mol} + \underbrace {HN{O_3}}_{0,28\,\,mol} \to \underbrace {NO}_{0,06\,\,mol} + \underbrace {\left\{ \begin{gathered} Al{(N{O_3})_3}:\,\,0,06\,\,mol \hfill \\ HN{O_3}:\,\,0,28 - 0,06.4 = 0,04\,\,mol \hfill \\ \end{gathered} \right\}}_{dd\,\,X}\)

\(+ \,\,\underbrace {Na}_{0,25} + HCl \to \underbrace {\left\{ \begin{gathered} NaCl:\,\,x\,\,mol \hfill \\ NaOH:\,\,y\,\,mol \hfill \\ \end{gathered} \right\}}_{dd\,\,Y}\)

\( + \,\,\underbrace {\left\{ \begin{gathered} Al{(N{O_3})_3}:\,\,0,06 \hfill \\ HN{O_3}:\,\,0,04 \hfill \\ \end{gathered} \right\}}_{dd\,\,X} + \underbrace {\left\{ \begin{gathered} NaCl:\,\,x \hfill \\ NaOH:\,\,y \hfill \\ \end{gathered} \right\}}_{dd\,\,Y} \to 0,02\,\,mol\,\,Al{(OH)_3} \downarrow \)

TH1: Al(OH)₃ không bị hòa tan

\( \Rightarrow \left\{ \begin{gathered} y = {n_{O{H^ - }}} = {n_{{H^ + }}} + 3{n_{Al{{(OH)}_3}}} = 0,1 \hfill \\ {n_{Na}} = x + y = 0,25 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} y = 0,1 \hfill \\ x = 0,15 \hfill \\ \end{gathered} \right. \Rightarrow [HCl] = \frac{{0,15}}{{0,5}} = \boxed{0,3M}\)

TH2: Al(OH)₃  bị hòa tan một phần

\(\Rightarrow \left\{ \begin{gathered} y = {n_{O{H^ - }}} = {n_{{H^ + }}} + 3{n_{A{l^{3 + }}}} + ({n_{A{l^{3 + }}}} - {n_{Al{{(OH)}_3}}}) = 0,26 \hfill \\ {n_{Na}} = x + y = 0,25 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} y = 0,26 \hfill \\ x = - 0,01 \hfill \\ \end{gathered} \right. \Rightarrow Vo\,\,li\)