`nFe = 2,8:56=0,05 (mol)`

`Fe + H2SO4 -> FeSO4 + H2↑`

`a, nH2= nFe = 0,05 (mol)`

` => VH2= 0,05.22,4=1,12 (l)`

`b, nFeSO4 =0,05 (mol)`

`mBaCl2=100.20,8%=20,8(g)`

`  => nBaCl2=0,1 (mol)`

 `BaCl2 + FeSO4 -> BaSO4 + FeCl2`

` nFeSO4 < nBaCl2 (0,05<0,1)`

` => FeSO4 hết ; BaCl2 dư`

` nBaSO4= nFeSO4 = 0,05 (mol)`

`  => mBaSO4=11,65 (g)`

`m dd = mFe + mddH2SO4 - mH2 + mddBaCl2`

        `  = 2,8 + 150 - 0,05.2 + 100`

        `  =252,7`

` nFeCl2 = nFeSO4 = 0,05 (mol)`

`  => mFeCl2=6,35 (g)`

 `  =>%FeCl2=6,35:252,7.100=2,5%`

`n BaCl2 pứ = nFeSO4=0,05 (mol)`

`=> nBaCl2 dư= 0,1-0,05=0,05 (mol)`

  `=> mBaCl2=10,4 (g)`

  ` => % BaCl2= 10,4:252,7.100=4,1%`

Đáp án:

 \(V=1,12 lít\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

Ta có:

\({n_{Fe}} = \frac{{2,8}}{{56}} = 0,05{\text{ mol = }}{{\text{n}}_{FeS{O_4}}} = {n_{{H_2}}}\)

\( \to V = {V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

BTKL:

\({m_{Fe}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to 2,8 + 150 = {m_{dd}} + 0,05.2\)

\( \to {m_{dd}} = 152,7{\text{ gam}}\)

Thêm vào 100 gam dung dịch \(BaCl_2\)

\(BaC{l_2} + FeS{O_4}\xrightarrow{{}}FeC{l_2} + BaS{O_4}\)

\({m_{BaC{l_2}}} = 100.20,8\%  = 20,8{\text{ gam}}\)

\( \to {n_{BaC{l_2}}} = \frac{{20,8}}{{137 + 35,5.2}} = 0,1{\text{ mol > }}{{\text{n}}_{FeS{O_4}}}\)

\( \to {n_{FeC{l_2}}} = {n_{BaS{O_4}}} = {n_{FeS{O_4}}} = 0,05{\text{ mol}}\)

\({n_{BaC{l_2}{\text{ dư}}}} = 0,1 - 0,05 = 0,05{\text{ mol}}\)

BTKL:

\({m_{dd}} + {m_{dd{\text{ BaC}}{{\text{l}}_2}}} = {m_{dd{\text{ sau phản ứng }}}} + {m_{BaS{O_4}}}\)

\( \to 152,7 + 100 = {m_{dd{\text{ sau phản ứng}}}} + 0,05.233\)

\( \to {m_{dd{\text{ sau phản ứng}}}} = 241,05{\text{ gam}}\)

\({m_{FeC{l_2}}} = 0,05.(56 + 35,5.2) = 6,35{\text{ gam}}\)

\({m_{BaC{l_2}}} = 0,05.(137 + 35,5.2) = 10,4{\text{ gam}}\)

\( \to C{\% _{FeC{l_2}}} = \frac{{6,35}}{{241,05}} = 2,63\% \)

\(C{\% _{BaC{l_2}}} = \frac{{10,4}}{{241,05}} = 4,31\% \)