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:

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

\( \to {V_{dd\;{\text{HCl}}}} = \frac{{0,2}}{3} = 0,067{\text{ lít}}\)

Mg+2HCl→MgCl2+H2

nMg=2,4/24=0,1mol

→nH2=0,1mol

VH2=0,1.22,4=2,24l

H2+ZnO→Zn+H2O

nZn=0,1mol

mZn=a=0,1.65=6,5g

nHCl=0,1.2=0,2mol

Vdd HCl=0,2/3=0,06l