Đáp án:
a) 6,72l
b) 150g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
nAl = \dfrac{m}{M} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
n{H_2} = \dfrac{{0,2 \times 3}}{2} = 0,3\,mol\\
V{H_2} = n \times 22,4 = 0,3 \times 22,4 = 6,72l\\
b)\\
n{H_2}S{O_4} = n{H_2} = 0,3\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,3 \times 98}}{{19,6\% }} = 150g\\
c)\\
nS{O_3} = \dfrac{{16}}{{80}} = 0,2\,mol\\
n{H_2}S{O_4} = nS{O_3} = 0,2\,mol\\
m{\rm{dd}} = 16 + 150 = 166g\\
C\% {H_2}S{O_4} = \dfrac{{0,2 \times 98}}{{166}} \times 100\% = 11,8\% \\
C\% A{l_2}{(S{O_4})_3} = \dfrac{{0,1 \times 342}}{{166}} \times 100\% = 20,6\%
\end{array}\)
