Đáp án:

\( {m_{Mg}}  = 2,4{\text{ gam}}; {{\text{m}}_{MgO}} = 4{\text{ gam}}\)

\({V_{HCl}} = 0,44{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

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

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

Ta có:

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

\( \to {m_{Mg}} = 0,1.24 = 2,4{\text{ gam}} \\\to {{\text{m}}_{MgO}} = 4{\text{ gam}}\)

\({n_{MgO}} = \dfrac{4}{{24 + 16}} = 0,1{\text{ mol}}\)

\( \to {n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,1.2 + 0,1.2 = 0,4{\text{ mol}}\)

\( \to {n_{HCl{\text{ tham  gia}}}} = 0,4.(100\%  + 10\% ) = 0,44{\text{ mol}}\)

\( \to {V_{HCl}} = \dfrac{{0,44}}{1} = 0,44{\text{ lít}}\)