Đáp án:

\({m_{Fe}} = 8,4{\text{ gam}}\)

\({C_{M{\text{ HCl}}}} = 1,5M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

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

Ta có: 

\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol = }}{{\text{n}}_{Fe}}\)

\( \to {m_{Fe}} = 0,15.56 = 8,4{\text{ gam}}\)

\({n_{HCl}} = 2{n_{Fe}} = 0,15.2 = 0,3{\text{ mol}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{0,3}}{{0,2}} = 1,5M\)

\({m_{HCl}} = 0,3.36,5 = 10,95{\text{ gam}}\)

\(\to {m_{dd{\text{ HCl}}}}{\text{ = }}\frac{{10,95}}{{3,65\% }}{\text{ = 300  gam}}\)

BTKL: 

\({m_{dd{\text{ muối}}}} = {m_{Fe}} + {m_{dd{\text{ HCl}}}} - {m_{{H_2}}}\)

\( = 8,4 + 300 - 0,15.2 = 308,1{\text{ gam}}\)

\({n_{FeC{l_2}}} = {n_{Fe}} = 0,15{\text{ mol}}\)

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

\( \to C{\% _{FeC{l_2}}} = \frac{{19,05}}{{308,1}} = 6,183\% \)

`n_(H_2)=\frac{3,36}{22,4}=0,15(mol)`

`a,`

`Fe+2HCl->FeCl_2+H_2`

`0,15`       `0,3`       `0,15`     `0,15`

`b,`

`m_(Fe)=0,15.56=8,4(g)`

`c,`

`C_(MHCl)=\frac{0,3}{0,2}=1,5M`

`d,`

`m_(dd)=8,4+\frac{0,3.36,5}{3,65%}-0,15.2=308,1(g)`

`m_(FeCl_2)=0,15.127=19,05(g)`

`C%_(FeCl_2)=\frac{19,05}{308,1}.100=6,183%`