Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
{n_{Al}} = \dfrac{{4,05}}{{27}} = 0,15mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{98 \times 14,7}}{{100}} = 14,406\\
{n_{{H_2}S{O_4}}} = \dfrac{{14,406}}{{98}} = 0,147mol\\
\dfrac{{0,15}}{2} > \dfrac{{0,147}}{3} \Rightarrow Al\text{ dư}\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{{H_2}S{O_4}}}}}{3} = 0,049mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,049 \times 342 = 16,758\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,147mol\\
{V_{{H_2}}} = 0,147 \times 22,4 = 3,2928l\\
c)\\
\text{Vì $H_2SO_4$ phản ứng hoàn toàn nên :}\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 98g\\
d)\\
{n_{A{l_d}}} = 0,15 - \dfrac{{0,147 \times 2}}{3} = 0,052g\\
{m_{A{l_d}}} = 0,052 \times 27 = 1,404g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{16,758}}{{98 + 4,05 - 1,404 - 0,147 \times 2}} \times 100\% = 16,7\%
\end{array}\)
