Đáp án:

\(\% {m_{Fe}} = 70\%;\% {m_{F{e_2}{O_3}}} = 30\% \)

\({C_{M{\text{ FeC}}{{\text{l}}_2}}}= 0,8834M;{C_{M{\text{ FeC}}{{\text{l}}_3}}} = 0,265M\)

\({C_{M{\text{ HCl}}}} = \frac{{0,0512}}{{0,1132}} = 0,4523M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

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

\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)

Ta có:

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Fe}}\)

\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}} \to {m_{F{e_2}{O_3}}} = 8 - 5,6 = 2,4{\text{ gam}}\)

\( \to \% {m_{Fe}} = \frac{{5,6}}{8} = 70\%  \to \% {m_{F{e_2}{O_3}}} = 30\% \)

Ta có:

\({n_{FeC{l_2}}} = {n_{Fe}} = 0,1;{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 2.\frac{{2,4}}{{160}} = 0,03{\text{ mol}}\)

Ta có:

\({n_{HCl{\text{ phản ứng}}}} = 2{n_{FeC{l_2}}} + 3{n_{FeC{l_3}}} = 0,1.2 + 0,03.3 = 0,29{\text{ mol}}\)

\( \to {n_{HCl{\text{ tham gia}}}} = \frac{{0,29}}{{100\%  - 15\% }} = 0,3412{\text{ mol}}\)

\({n_{HCl{\text{ dư}}}} = 0,3412 - 0,29 = 0,0512{\text{ mol}}\)

\({m_{HCl{\text{ tham gia}}}} = 0,3412.36,5 = 12,4538{\text{ gam}}\)

\( \to {m_{dd{\text{ HCl}}}} = \frac{{12,4538}}{{10\% }} = 124,538{\text{ gam}} \to {{\text{V}}_{dd\;{\text{HCl}}}} = \frac{{124,538}}{{1,1}} = 113,2{\text{ ml = 0}}{\text{,1132 lít}}\)

\( \to {C_{M{\text{ FeC}}{{\text{l}}_2}}} = \frac{{0,1}}{{0,1132}} = 0,8834M;{C_{M{\text{ FeC}}{{\text{l}}_3}}} = \frac{{0,03}}{{0,1132}} = 0,265M\)

\({C_{M{\text{ HCl}}}} = \frac{{0,0512}}{{0,1132}} = 0,4523M\)

Giải thích các bước giải:

`a.`

`-n_{H_2}=\frac{2,24}{22,4}=0,1(mol)`

`→m_{H_2}=0,1.2=0,2(g)`

Phương trình hóa học :

 `Fe+2HCl→FeCl_2+H_2↑`

`0,1`        `0,2`             `0,1` `←`   `0,1`             `(mol)`

 `Fe_2O_3+6HCl→2FeCl_3+H_2O`

`0,015`   `→`   `0,09`          `0,03`                      `(mol)`

`-m_{Fe}=0,1.56=5,6(g)`

`-m_{Fe_2O_3}=m_{hh}-m_{Fe}=8-5,6=2,4(g)`

`→%m_{Fe}=\frac{5,6}{8}.100%=70%`

`→%m_{Fe_2O_3}=100%-%m_{Fe}=100%-70%=30%`

`b.`

`-n_{Fe_2O_3}=\frac{2,4}{160}=0,015(mol)`

`-∑n_{HCl}=0,2+0,09=0,29(mol)`

Vì dùng dư `15%` dung dịch `HCl`

`→n_{HCl}` cần dùng `=0,29.\frac{100%}{100%-15%}=\frac{29}{85}(mol)`

`→m_{ct.HCl}=\frac{29}{85}.36,5≈12,453(g)`

`→m_{dd.HCl}=\frac{12,453.100%}{10%}=124,53(g)`

`-n_{HCl}` dư `=\frac{29}{85}-0,29=\frac{87}{1700}(mol)`

Dung dịch sau phản ứng gồm :

`FeCl_2:0,1(mol)`

`FeCl_3:0,03(mol)`

`HCl` dư `:\frac{87}{1700}(mol)`

`-m_{dd.sau.pứ}=m_{hh}+m_{dd.HCl}-m_{H_2}`

`⇔m_{dd.sau.pứ}=8+124,53-0,2=132,33(g)`

`-m_{ct.FeCl_2}=0,1.127=12,7(g)`

`→C%FeCl_2=\frac{12,7}{132,33}.100%≈9,6%`

`-m_{ct.FeCl_3}=0,03.162,5=4,875(g)`

`→C%FeCl_3=\frac{4,875}{132,33}.100%≈3,684%`

`-m_{ct.HCl}` dư `=\frac{87}{1700}.36,5≈1,868(g)`

`→C%HCl` dư `=\frac{1,868}{132,33}.100%≈1,41%`

Vì `d_{HCl}=1,1`$(g/ml)$

`→V_{HCl}=\frac{124,53}{1,1}≈113,21(ml)=0,11321(l)`

Mà thể tích dung dịch thay đổi không đáng kể

`-V_{dd.sau.pứ}=V_{dd.ban.đầu}=V_{HCl}=0,11321(l)`

`→CM_{FeCl_2}=\frac{0,1}{0,11321}≈0,883(M)`

`→CM_{FeCl_3}=\frac{0,03}{0,11321}≈0,265(M)`

`→CM_{HCl}` dư `=\frac{\frac{87}{1700}}{0,11321}≈0,452(M)`

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