Câu 1.

${n_{KOH}} = 0,5.3,75 = 1,875mol \to {n_{O{H^ - }}} = 1,875mol$

${n_{A{l_2}{{(S{O_4})}_3}}} = 0,25mol \to {n_{A{l^{3 + }}}} = 0,5mol $

$A{l^{3 + }} + 3O{H^ - } \to Al{(OH)_3}$

$0,5$ $→$  $1,5$        $0,5$

$O{H^ - } + Al{(OH)_3} \to AlO_2^ -  + 2{H_2}O$

$0,375$ $→$ $0,375$

$ \Rightarrow {n_{Al{{(OH)}_3}}} = 0,5 - 0,375 = 0,125mol$

$ \Rightarrow m = 0,125.78 = 9,75g$

Câu 2.

$\begin{gathered}   {n_{HCl}} = 0,02mol;{n_{{H_2}S{O_4}}} = 0,01mol \hfill \\    \Rightarrow {n_{{H^ + }}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 0,04mol \hfill \\ \end{gathered} $

${n_{Ba{{(OH)}_2}}} = 0,3a \Rightarrow {n_{O{H^ - }}} = 0,6a$

$pH = 13 → O{H^ - }$ dư

$→ pOH = 14 – 13 = 1$ $ \Rightarrow {\left[ {O{H^ - }} \right]_{dư}} = 0,1M \Rightarrow {n_{O{H^ - }(dư)}} = 0,1.0,5 = 0,05\,\,mol$

PTHH: ${H^ + } + O{H^ - } \to {H_2}O$

Bd:  $0,04$    $0,6a$

Sau:   $0$     $0,6a-0,04$

$ \Rightarrow 0,6a - 0,04 = 0,05 \Rightarrow a = 1,5$

$ \Rightarrow {n_{Ba{{(OH)}_2}}} = 0,45\,\,mol \Rightarrow {n_{B{a^{2 + }}}} = 0,45\,\,mol$

$B{a^{2 + }} + SO_4^{2 - } \to BaS{O_4}$

$0,45$    $0,01$

$ \Rightarrow {n_{BaS{O_4}}} = 0,01\,\,mol \Rightarrow {m_{BaS{O_4}}} = 2,33\,\,g$