Đáp án:

$V=5,04(l)$

$m=9,45(g)$ 

Giải thích các bước giải:

$n_{Al}=\frac{4,05}{27}=0,15(mol)$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

Theo PT: $n_{H_2}=1,5n_{Al}=0,225(mol)$

$\to V_{H_2}=0,225.22,4=5,04(l)$

$Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O$

Theo PT: $n_{Fe}=\frac{3}{4}n_{H_2}=0,16875(mol)$

$\to m_{Fe}=0,16875.56=9,45(g)$

Em tham khảo nha :

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{4,05}}{{27}} = 0,15mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,225mol\\
{V_{{H_2}}} = 0,225 \times 22,4 = 5,04l\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
{n_{Fe}} = \dfrac{3}{4}{n_{{H_2}}} = 0,16875mol\\
{m_{Fe}} = 0,16875 \times 56 = 9,45g
\end{array}\)