Đáp án:
a)
\(m = 42,75g\)
b)
\({V_{Ba{{(OH)}_2}}} = 1,25ml\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
A{l_2}{(S{O_4})_3} + 3Ba{(OH)_2} \to 2Al{(OH)_3} + 3BaS{O_4}\\
{n_{Ba{{(OH)}_2}}} = 0,35 \times 0,5 = 0,175mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = 01 \times 0,5 = 0,05mol\\
\dfrac{{0,175}}{3} > \dfrac{{0,05}}{1} \Rightarrow Ba{(OH)_2}\text{ dư}\\
{n_{Al{{(OH)}_3}}} = 2 \times {n_{A{l_2}{{(S{O_4})}_3}}} = 2 \times 0,05 = 0,1mol\\
{m_{Al{{(OH)}_3}}} = 0,1 \times 78 = 7,8g\\
{n_{BaS{O_4}}} = 3 \times {n_{A{l_2}{{(S{O_4})}_3}}} = 3 \times 0,05 = 0,15mol\\
{m_{BaS{O_4}}} = 0,15 \times 233 = 34,95g\\
m = {m_{BaS{O_4}}} + {m_{Al{{(OH)}_3}}} = 7,8 + 34,95 = 42,75g\\
2)\\
pOH = 14 - 12 = 2\\
\Rightarrow {\rm{[}}O{H^ - }{\rm{]}} = 0,01M\\
{n_{O{H^ - }}} = 0,1 \times 0,01 = 0,001mol\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{0,001}}{2} = 0,0005mol\\
{V_{Ba{{(OH)}_2}}} = \dfrac{{0,0005}}{{0,4}} = 0,00125l = 1,25ml
\end{array}\)
