Em tham khảo nha :

\(\begin{array}{l}
2)\\
a)\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
{n_{{H_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45mol\\
{n_{Fe}} = \dfrac{3}{4}{n_{{H_2}}} = 0,3375mol\\
{m_{Fe}} = 0,3375 \times 56 = 18,9g\\
b)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{HCl}} = \dfrac{{10,95}}{{36,5}} = 0,3mol\\
\dfrac{{0,3375}}{1} > \dfrac{{0,3}}{2} \Rightarrow Fe\text{ dư}\\
{n_{F{e_d}}} = 0,3375 - \dfrac{{0,3}}{2} = 0,1875mol\\
{m_{F{e_d}}} = 0,1875 \times 56 = 10,5g\\
{n_{FeC{l_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,15mol\\
{m_{FeC{l_2}}} = 0,15 \times 127 = 19,05\\
3)\\
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{4,05}}{{27}} = 0,15mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,225mol\\
{V_{{H_2}}} = 0,225 \times 22,4 = 5,04l\\
b)\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
{n_{F{e_3}{O_4}}} = \dfrac{{23,2}}{{232}} = 0,1mol\\
\dfrac{{0,1}}{1} > \dfrac{{0,225}}{4} \Rightarrow F{e_3}{O_4}\text{ dư}\\
{n_{F{e_3}{O_4}d}} = 0,1 - \dfrac{{0,225}}{4} = 0,04375mol\\
{m_{F{e_3}{O_4}d}} = 0,04375 \times 232 = 10,15g\\
{n_{Fe}} = \dfrac{3}{4}{n_{{H_2}}} = 0,16875mol\\
{m_{Fe}} = 0,16875 \times 56 = 9,45g
\end{array}\)

Bài $1,$

Phản ứng xảy ra:

\(F{e_3}{O_4} + 4{H_2}\xrightarrow{{{t^o}}}3Fe + 4{H_2}O\)

Ta có:

\({n_{{H_2}}} = \frac{{10,08}}{{22,4}} = 0,45{\text{ mol}}\)

\( \to {n_{Fe}} = \frac{3}{4}{n_{{H_2}}} = \frac{3}{4}.0,45 = 0,3375{\text{ mol}}\)

\( \to {m_{Fe}} = 0,3375.56 = 18,9{\text{ gam}}\)

Cho lượng \(Fe\) trên tác dụng với \(HCl\)

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Ta có:

\({n_{HCl}} = \frac{{10,95}}{{36,5}} = 0,3{\text{ mol}}\)

\( \to {n_{Fe}} > \frac{1}{2}{n_{HCl}}\) nên \(Fe\) dư

\( \to {n_{Fe{\text{ dư}}}} = 0,3375 - \frac{{0,3}}{2} = 0,1875{\text{ mol}}\)

\({n_{FeC{l_2}}} = \frac{1}{2}{n_{HCl}} = 0,15{\text{ mol}}\)

\( \to {m_{Fe{\text{ dư}}}} = 0,1875.56 = 10,5{\text{ gam}}\)

\({m_{FeC{l_2}}} = 0,15.(56 + 35,5.2) = 19,05{\text{ gam}}\)

Bài $2,$

$a,$

`nAl= frac{4,05}{27} =0,15` (mol)

Theo phương trình: `nH2 =0,225` (mol)

`→ VH2(đktc)=0,225×22,4=5,04 (l)`

$b,$ 

nFe3O4 = 23,2232 = 0,1 (mol)

Xét TLD giữa Fe3O4 và H2, ta có:

0,11 < 0,2254

`→` Fe3O4 dư, H2 hết.

`→` Xét theo (mol) của H2.

`→` mFe3O4 (dư) = `(0,1−0,05625) × 232 = 10,15` (g)

Theo phương trình nFenFe =0,16875(mol)

`→` mFemFe = `0,16875 × 56 = 9,45` (g)