Đáp án:
\({{\text{m}}_{Zn}} = 48,75{\text{ gam}}\)
\({{\text{m}}_{dd{\text{ HCl}}}} = 375{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = {n_{Zn}} = \frac{{16,8}}{{22,4}} = 0,75{\text{ mol}} \to {{\text{m}}_{Zn}} = 0,75.65 = 48,75{\text{ gam}}\)
Ta có:
\({n_{HCl}} = 2{n_{{H_2}}} = 1,5{\text{ mol}} \to {{\text{m}}_{HCl}} = 1,5.36,5 = 54,75{\text{ gam}} \to {{\text{m}}_{dd{\text{ HCl}}}} = \frac{{54,75}}{{14,6\% }} = 375{\text{ gam}}\)
BTKL:
\({m_{dd{\text{ sau phản ứng}}}} = {m_{Zn}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} - {m_{{H_2}}} = 48,75 + 375 - 0,75.2 = 422,25{\text{ gam}}\)
Ta có:
\({n_{ZnC{L_2}}} = {n_{Zn}} = 0,75{\text{ mol}} \to {{\text{m}}_{ZnC{l_2}}} = 0,75.(65 + 35,5.2) = 102{\text{ gam}}\)
\( \to C\% {m_{ZnC{l_2}}} = \frac{{102}}{{422,25}} = 24,16\% \)
