Đáp án:
\( {V_{{H_2}}} = 2,24{\text{ lít}}\)
\({C_{M{\text{ HCl}}}} = 0,5M\)
\( {m_{ZnC{l_2}}} = 13,6{\text{ gam}}\)
\( {m_{Hg}}= 20,1{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({n_{HCl}} = 2{n_{Zn}} = 0,2{\text{ mol}}\)
\({C_{M{\text{ HCl}}}} = \frac{{0,2}}{{0,4}} = 0,5M\)
\({n_{ZnC{l_2}}} = {n_{Zn}} = 0,1{\text{ mol}}\)
\( \to {m_{ZnC{l_2}}} = 0,1.(65 + 35,5.2) = 13,6{\text{ gam}}\)
Dùng lượng \(H_2\) trên khử \(HgO\)
\(HgO + {H_2}\xrightarrow{{{t^o}}}Hg + {H_2}O\)
\({n_{Hg}} = {n_{{H_2}}} = 0,1{\text{ mol}}\)
\( \to {m_{Hg}} = 0,1.201 = 20,1{\text{ gam}}\)