Đáp án:
\( {V_{{H_2}}} = 2,24{\text{ lít}}\)
\( C{\% _{ZnC{l_2}}} = 12,79\% ;C{\% _{HCl}} = 6,87\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol}}\)
\({m_{HCl}} = 100.14,6\% = 14,6{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{14,6}}{{36,5}} = 0,4{\text{ mol > 2}}{{\text{n}}_{Zn}}\)
Vậy \(HCl\) dư
\( \to {n_{{H_2}}} = {n_{ZnC{l_2}}} = {n_{Zn}} = 0,1{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({n_{HCl{\text{ dư}}}} = 0,4 - 0,1.2 = 0,2{\text{ mol}}\)
BTKL:
\({m_{Zn}} + {m_{dd{\text{ HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 6,5 + 100 = {m_{dd}} + 0,1.2 \to {m_{dd}} = 106,3{\text{ gam}}\)
\({m_{ZnC{l_2}}} = 0,1.(65 + 35,5.2) = 13,6{\text{ gam}}\)
\({m_{HCl{\text{ dư}}}} = 0,2.36,5 = 7,3{\text{ gam}}\)
\( \to C{\% _{ZnC{l_2}}} = \frac{{13,6}}{{106,3}} = 12,79\% ;C{\% _{HCl}} = \frac{{7,3}}{{106,3}} = 6,87\% \)
