Đáp án:

 a) %mMg=23%; %mMgO=77%

b) m dung dịch HCl=150 gam

c) C% MgCl2=9,19%

Giải thích các bước giải:

 Phản ứng xảy ra:

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

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

\(\to {n_{Mg}} = {n_{{H_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol}} \to {{\text{m}}_{Mg}} = 0,05.24 = 1,2{\text{ gam}} \to {{\text{m}}_{MgO}} = 5,2 - 1,2 = 4{\text{ gam}} \to {{\text{n}}_{MgO}} = \frac{4}{{24 + 16}} = 0,1{\text{ mol}}\)

\(\to \% {m_{Mg}} = \frac{{1,2}}{{5,2}} = 23\%  \to \% {m_{MgO}} = 77\% \)

\({n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,05.2 + 0,1.2 = 0,3{\text{ mol}}\)

\(\to {m_{HCl}} = 0,3.36,5 = 10,95{\text{ gam}} \to {{\text{m}}_{dd{\text{ HCl}}}} = \frac{{10,95}}{{7,3\% }} = 150{\text{ gam}}\)

BTKL: \({m_{dd{\text{ sau phản ứng}}}} = 5,2 + 150 - 0,05.2 = 155,1{\text{ gam}}\)

Dung dịch sau phản ứng chứa MgCl2.

\({n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,15{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 0,15(24 + 35,5.2) = 14,25{\text{ gam}}\)

\(\to C{\% _{MgC{l_2}}} = \frac{{14,25}}{{155,1}} = 9,19\% \)