Em tham khảo nha :
\(\begin{array}{l}
3)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,1 \times 0,5 = 0,05mol\\
\dfrac{{0,2}}{2} > \dfrac{{0,05}}{3} \Rightarrow Al\text{ dư}\\
a)\\
C\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,05mol\\
{V_{{H_2}}} = 0,05 \times 22,4 = 1,12l\\
b)\\
A\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{{H_2}S{O_4}}}}}{3} = \dfrac{1}{{60}}mol\\
{C_{{M_{A{l_2}{{(S{O_4})}_3}}}}} = \dfrac{n}{V} = \dfrac{1}{{60}}:0,1 = 0,17M
\end{array}\)
