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$
