Đáp án:

9. 10,8g

10. 18g 

Giải thích các bước giải:

Bài 9: 

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
{n_{F{e_3}{O_4}}} = 0,15mol\\
 \to {n_{{H_2}}} = 4{n_{F{e_3}{O_4}}} = 0,6mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,4mol\\
 \to {m_{Al}} = 10,8g
\end{array}\)

Bài 10:

\(\begin{array}{l}
2{H_2} + {O_2} \to 2{H_2}O\\
{n_{{H_2}}} = 1mol\\
{n_{{O_2}}} = 0,625mol\\
 \to \dfrac{{{n_{{H_2}}}}}{2} < \dfrac{{{n_{{O_2}}}}}{1} \to {n_{{O_2}}}dư\\
 \to {n_{{H_2}O}} = {n_{{H_2}}} = 1mol\\
 \to {m_{{H_2}O}} = 18g
\end{array}\)

Bài 9

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`Fe_3O_4+4H_2->3Fe+4H_2O`

`n_(Fe_3O_4)=(34,8)/(56.3+16.4)=0,15(mol)`

`=>n_(H_2)=4.0,15=0,6(mol)`

`=>n_(Al)=2/3 *0,6=0,4(mol)`

`m_(Al)=27.0,4=10,8(g)`

Bài 10

`2H_2+O_2->2H_2O`

`n_(H_2)=1(mol)`

`n_(O_2)=0,625(mol)`

`=>O_2` dư.

`=>n_(H_2O)=n_(H_2)=1(mol)`

`m_(H_2O)=18.1=18(g)`