Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
6NaOH + A{l_2}{(S{O_4})_3} \to 3N{a_2}S{O_4} + 2Al{(OH)_3}\\
{n_{Al{{(OH)}_3}}} = 0,15mol\\
 \to {n_{NaOH}} = 3{n_{Al{{(OH)}_3}}} = 0,45mol\\
 \to {n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al{{(OH)}_3}}} = 0,075mol\\
 \to C{M_{NaOH}} = \dfrac{{0,45}}{{0,2}} = 2,25M\\
 \to C{M_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,075}}{{0,3}} = 0,25M\\
b,\\
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
{n_{A{l_2}{O_3}}} = \dfrac{1}{2}{n_{Al{{(OH)}_3}}} = 0,075mol\\
 \to {m_{A{l_2}{O_3}}} = 7,65g
\end{array}\)

Bạn tham khảo:

$a/$

$Al_2(SO_4)_3+6NaOH \to 2Al(OH)_3+3Na_2SO_4$

$n_{Al(OH)_3}=\dfrac{11,7}{78}=0,15(mol)$

$n_{Al_2(SO_4)_3}=\dfrac{0,15}{2}=0,075(mol)$

$CM_{Al_2(SO_4)_3}=\dfrac{0,075}{0,3}=0,25M$

$n_{NaOH}=0,15.3=0,45(mol)$

$CM_{NaOH}=\dfrac{0,45}{0,2}=2,25M$
$b/$

$2Al(OH)_3 \xrightarrow{t^{o}} Al_2O_3+3H_2O$

$n_{Al_2O_3}=\dfrac{0,15}{2}=0,075(mol)$

$m_{Al_2O_3}=0,075.102=7,65(g)$