Đáp án:

 a) 3,36l

b) 17,1 g và 34,3g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
nAl = \dfrac{m}{M} = \dfrac{{2,7}}{{27}} = 0,1\,mol\\
n{H_2}S{O_4} = \dfrac{{200 \times 24,5\% }}{{98}} = 0,5\,mol\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
\dfrac{{0,1}}{2} < \dfrac{{0,5}}{3} \Rightarrow\text{$H_2SO_4$ dư} \\
n{H_2} = 0,1 \times 1,5 = 0,15\,mol\\
V{H_2} = n \times 22,4 = 0,15 \times 22,4 = 3,36l\\
b)\\
n{H_2}S{O_4} = 0,5 - 0,15 = 0,35\,mol\\
nA{l_2}{(S{O_4})_3} = \dfrac{{0,1}}{2} = 0,05\,mol\\
m{H_2}S{O_4} = n \times M = 0,35 \times 98 = 34,3g\\
mA{l_2}{(S{O_4})_3} = n \times M = 0,05 \times 342 = 17,1g\\
c)\\
mddspu = 2,7 + 200 - 0,15 \times 2 = 202,4g\\
C\% {H_2}S{O_4} = \dfrac{{mct}}{{mdd}} \times 100\%  = \dfrac{{34,3}}{{202,4}} \times 100\%  = 16,95\% \\
C\% A{l_2}{(S{O_4})_3} = \dfrac{{mct}}{{mdd}} \times 100\%  = \dfrac{{17,1}}{{202,4}} \times 100\%  = 8,45\% 
\end{array}\)