Đáp án:

$\downarrow$

Giải thích các bước giải:

* `n_{Al}=(5,4)/27=0,2(mol)`

a, `-` PTHH:    `2Al+3H_2SO_4` $\xrightarrow{t^o}$ `Al_2(SO_4)_3+3H_2`

`->n_{H_2}=3/2 n_{Al}=0,3(mol)`

`->V_{H_2(đktc)}=0,3xx22,4=6,72(lít)`

b, * Theo PTHH `->n_{H_2SO_4}=n_{H_2}=0,3(mol)`

`->m_{H_2SO_4}=0,3xx98=29,4(gam)`

c, * Ta thấy: `n_{Al_2(SO_4)_3}=1/2 n_{Al}=0,1(mol)`

`->m_{Al_2(SO_4)_3}=0,1xx342=34,2(gam)`

$\text{#Kudo}$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{m_{{H_2}S{O_4}}} = 0,3 \times 98 = 29,4g\\
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,1 \times 342 = 34,2g
\end{array}\)