Cho 8,905 gam Ba tan hết vào V ml dung dịch Al

A

- Hướng dẫn giải

\({n_{Ba}} = 0,065\xrightarrow{{}}{m_ \downarrow } = 16,32\xrightarrow{{}}\left\{ \begin{gathered} BaS{O_4}:a \hfill \\ Al{(OH)_3}:b \hfill \\ B{a^{2 + }}:0,065 - a \hfill \\ AlO_2^ - :0,13 - 2a \hfill \\ \end{gathered} \right.\)

\(\xrightarrow{{}}\left\{ \begin{gathered} 233a + 78b = 16,32 \hfill \\ a = 1,5(b + 0,13 - 2a) \hfill \\ \end{gathered} \right.\xrightarrow{{}}\left\{ \begin{gathered} a = 0,06 \hfill \\ b = 0,03 \hfill \\ \end{gathered} \right.\xrightarrow[{}]{}V = 200(ml)\)