Đáp án:
\(\begin{array}{l}
a.\\
{m_{A{l_2}{O_3}}} = 10,2g\\
{m_{{\rm{BaS}}{O_4}(*)}} = 69,9g\\
b.\\
C{M_{A{l_2}{{(S{O_4})}_3}(dư)}} = 0,4M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A{l_2}{(S{O_4})_3} + 3Ba{(OH)_2} \to 2Al{(OH)_3} + 3B{\rm{aS}}{O_4}\\
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O
\end{array}\)
\(\begin{array}{l}
{n_{A{l_2}{{(S{O_4})}_3}}} = 0,2mol\\
{n_{Ba{{(OH)}_2}}} = 0,3mol\\
\to {n_{A{l_2}{{(S{O_4})}_3}}} > \dfrac{{{n_{Ba{{(OH)}_2}}}}}{3} \to {n_{A{l_2}{{(S{O_4})}_3}}}dư
\end{array}\)
\(\begin{array}{l}
3BaC{l_2} + A{l_2}{(S{O_4})_3} \to 3B{\rm{aS}}{O_4} + 2AlC{l_3}\\
D:A{l_2}{O_3}\\
E:B{\rm{aS}}{O_4}
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{Al{{(OH)}_3}}} = \dfrac{2}{3}{n_{Ba{{(OH)}_2}}} = 0,2mol\\
\to {n_{A{l_2}{O_3}}} = \dfrac{1}{2}{n_{Al{{(OH)}_3}}} = 0,1mol\\
\to {m_{A{l_2}{O_3}}} = 10,2g
\end{array}\)
\(\begin{array}{l}
{n_{A{l_2}{{(S{O_4})}_3}(pt)}} = \dfrac{1}{3}{n_{Ba{{(OH)}_2}}} = 0,1mol\\
\to {n_{A{l_2}{{(S{O_4})}_3}(dư)}} = 0,1mol\\
\to {n_{{\rm{BaS}}{O_4}(*)}} = 3{n_{A{l_2}{{(S{O_4})}_3}(dư)}} = 0,3mol\\
\to {m_{{\rm{BaS}}{O_4}(*)}} = 69,9g
\end{array}\)
\(\begin{array}{l}
b.\\
C{M_{A{l_2}{{(S{O_4})}_3}(du)}} = \dfrac{{0,1}}{{0,05 + 0,2}} = 0,4M
\end{array}\)
