Đáp án:
\({m_1} = 2,6{\text{ gam;}}{{\text{m}}_2} = 20{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{0,896}}{{22,4}} = 0,04{\text{ mol = }}{{\text{n}}_{Zn}} \to {n_{HCl}} = 2{n_{Zn}} = 0,08{\text{ mol}}\)
Ta có:
\({m_1} = {m_{Zn}} = {n_{{H_2}}} = 0,04{\text{ mol}} \)
\(\to {{\text{m}}_1} = 0,04.65 = 2,6{\text{ gam}}\)
\({m_{HCl}} = 0,08.36,5 = 2,92{\text{ gam}} \to {{\text{m}}_2} = \frac{{2,92}}{{14,6\% }} = 20{\text{ gam}}\)
BTKL:
\({m_{dd{\text{ }}}} = {m_{Zn}} + {m_{dd{\text{HCl}}}} - {m_{{H_2}}} \)
\(= 2,6 + 20 - 0,04.2 = 32,52gam\)
\({n_{ZnC{l_2}}} = {n_{Zn}} = 0,04{\text{ mol}}\)
\( \to {{\text{m}}_{ZnC{l_2}}} = 0,04.(65 + 35,5.2) = 5,44{\text{ gam}}\)
\( \to C{\% _{ZnC{l_2}}} = \frac{{5,44}}{{32,52}} = 16,73\% \)
