Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3mol\\
{m_{{H_2}S{O_4}}} = 0,3 \times 98 = 29,4g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{29,4 \times 100}}{{19,6}} = 150g\\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,1 \times 342 = 34,2g\\
{m_{{\rm{dd}}spu}} = 5,4 + 150 - 0,3 \times 2 = 154,8g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{34,2}}{{154,8}} \times 100\% = 22,09\%
\end{array}\)
