Đáp án:
\(a\% = 3,65\% \)
\(C{\% _{ZnC{l_2}}} = 6,59\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol = }}{{\text{n}}_{ZnC{l_2}}} = {n_{{H_2}}}\)
\( \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\({n_{HCl}} = 2{n_{Zn}} = 0,4{\text{ mol}}\)
\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)
\( \to a\% = C{\% _{HCl}} = \frac{{14,6}}{{400}} = 3,65\% \)
BTKL:
\({m_{Zn}} + {m_{dd{\text{ HCl}}}} = {m_{dd\;{\text{A}}}} + {m_{{H_2}}}\)
\( \to 13 + 400 = {m_{dd{\text{ A}}}} + 0,2.2 \to {m_{dd\;{\text{A}}}} = 412,6{\text{ gam}}\)
\({m_{ZnC{l_2}}} = 0,2.(65 + 35,5.2) = 27,2{\text{ gam}}\)
\( \to C{\% _{ZnC{l_2}}} = \frac{{27,2}}{{412,6}} = 6,59\% \)
