Đáp án:
$a.\,\,{V_{{H_2}}} = 6,72\,\,lít$
$b.\,\,{m_{A{l_2}{{(S{O_4})}_3}}} = 34,2\,\,gam$
Giải thích các bước giải:
a.
${n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,\,mol$
Phương trình hóa học:
$2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}$
$\,\,0,2\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,3\,\,\,\,\,\,\,\,mol$
$ \to {V_{{H_2}}} = 0,3.22,4 = 6,72\,\,lít$
b.
Theo PTHH: ${n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,1\,\,mol$
$ \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,1.342 = 34,2\,\,gam$