Đáp án:
\( V = 2,24{\text{ lít}}\)
\(a = 6,5{\text{ gam}}\)
\({V_{dd\;{\text{HCl}}}} = 0,067{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
\(ZnO + {H_2}\xrightarrow{{{t^o}}}Zn + {H_2}O\)
Ta có:
\({n_{Mg}} = \frac{{2,4}}{{24}} = 0,1{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
\( \to {n_{{H_2}}} = {n_{Zn}} = 0,1{\text{ mol}}\)
\( \to V = {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\(a = {m_{Zn}} = 0,1.65 = 6,5{\text{ gam}}\)
Ta có:
\({n_{HCl}} = 2{n_{Mg}} = 0,2{\text{ mol}}\)
\( \to {V_{dd\;{\text{HCl}}}} = \frac{{0,2}}{3} = 0,067{\text{ lít}}\)
