Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 13,44l\\
b)\\
{C_\% }A{l_2}{(S{O_4})_3} = 17,03\% \\
c)\\
{m_{Al{{(OH)}_3}}} = 15,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{10,8}}{{27}} = 0,4mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{392 \times 15\% }}{{98}} = 0,6\,mol\\
\dfrac{{0,4}}{2} = \dfrac{{0,6}}{3} \Rightarrow\text{ Phản ứng vừa đủ} \\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,6\,mol\\
{V_{{H_2}}} = 0,6 \times 22,4 = 13,44l\\
b)\\
{m_{{\rm{dd}}spu}} = 10,8 + 392 - 0,6 \times 2 = 401,6g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,2 \times 342}}{{401,6}} \times 100\% = 17,03\% \\
c)\\
{n_{NaOH}} = 0,7 \times 2 = 1,4\,mol\\
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
\dfrac{{0,2}}{1} < \dfrac{{1,4}}{6} \Rightarrow NaOH\text{ dư}\\
{n_{NaOH}} \text{ dư}= 1,4 - 0,2 \times 6 = 0,2\,mol\\
{n_{Al{{(OH)}_3}}} = 2{n_{A{l_2}{{(S{O_4})}_3}}} = 0,2 \times 2 = 0,4\,mol\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
\dfrac{{0,4}}{1} > \dfrac{{0,2}}{1} \Rightarrow\text{ $Al(OH)_3$ dư} \\
{n_{Al{{(OH)}_3}}}\text{ dư} = 0,4 - 0,2 = 0,2\,mol\\
{m_{Al{{(OH)}_3}}} = 0,2 \times 78 = 15,6g
\end{array}\)
