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:
\({n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Mg}}\)
\( \to {m_{Mg}} = 0,1.24 = 2,4{\text{ gam}} \\\to {{\text{m}}_{MgO}} = 4{\text{ gam}}\)
\({n_{MgO}} = \dfrac{4}{{24 + 16}} = 0,1{\text{ mol}}\)
\( \to {n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,1.2 + 0,1.2 = 0,4{\text{ mol}}\)
\( \to {n_{HCl{\text{ tham gia}}}} = 0,4.(100\% + 10\% ) = 0,44{\text{ mol}}\)
\( \to {V_{HCl}} = \dfrac{{0,44}}{1} = 0,44{\text{ lít}}\)