Đáp án:

\( {V_{{H_2}}}= 2,24{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)

Ta có: 

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

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

Ta có: 

\({n_{HCl}} = 2{n_{Mg}} = 0,2{\text{ mol}}\)

\( \to {m_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}}\)

\( \to C{\% _{HCl}} = x\%  = \frac{{7,3}}{{200}} = 3,65\% \)

\({n_{MgC{l_2}}} = {n_{Mg}} = 0,1{\text{ mol}}\)

\( \to {m_{MgC{l_2}}} = 0,1.(24 + 35,5.2) = 9,5{\text{ gam}}\)

BTKL:

\({m_{dd{\text{ sau phản ứng}}}} = {m_{Mg}} + {m_{dd{\text{ HCl}}}} - {m_{{H_2}}}\)

\( = 2,4 + 200 - 0,1.2 = 202,2{\text{ gam}}\)

\( \to C{\% _{MgC{l_2}}} = \frac{{9,5}}{{202,2}} = 4,698\% \)

 `a)`

`Mg + 2HCl → MgCl_2 + H_2` 

`0,1 → 0,2 → 0,1 → 0,1(mol)`

`n_{HCl} = {2,4}/{24} = 0,1(mol)`

`V_{H_2} = 0,1 . 22,4 = 2,24(l)`

`b)`

`x = {0,2. 36,5}/{200} .100 = 3,65%`

`c)`

`m_{MgCl_2} = 0,1 . 95 = 9,5(g)`

`d)`

`mdd = 2,4 + 200 - 0,1. 2 = 202,2(g)`

`e)`

`C%_{MgCl_2} = {9,5}/{202,2} .100 = 4,7%`