Đáp án:

 a) \(\% {m_{Zn}} = 65\% ;{\text{ \% }}{{\text{m}}_{Cu}} = 35\% \)

b) x=80,3 gam

c) \(C{\% _{ZnC{l_2}}} = 22,73\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ du}}}} = 1,22{\text{\% }}\)

Giải thích các bước giải:

 Cu không phản ứng với HCl, chỉ có Zn phản ứng.

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

\(\to {n_{Zn}} = {n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \to {{\text{m}}_{Zn}} = 0,15.65 = 9,75{\text{ gam}}\)

\(\to \% {m_{Zn}} = \frac{{9,75}}{{15}} = 65\%  \to \% {m_{Cu}} = 35\% \)

Ta có: \({n_{HCl{\text{ phản ứng}}}} = 2{n_{{H_2}}} = 0,3{\text{ mol}} \to {{\text{n}}_{HCl{\text{ tham gia}}}} = 0,3.110\%  = 0,33{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,33.36,5 = 12,045{\text{ gam}}\)

\(\to x = \frac{{12,045}}{{15\% }} = 80,3{\text{ gam}}\)

c) BTKL:  \({m_Y} = 9,75 + 80,3 - 0,15.2 = 89,75{\text{ gam}}\)

Ta có: \({n_{ZnC{l_2}}} = 0,15{\text{ mol;}}\;{{\text{n}}_{HCl{\text{ dư}}}} = 0,3.10\%  = 0,03{\text{ mol}}\)

\({m_{ZnC{l_2}}} = 0,15.(65 + 35,5.2) = 20,4 \to C{\% _{ZnC{l_2}}} = \frac{{20,4}}{{89,75}} = 22,73\% \)

\({m_{HCl{\text{ dư}}}} = 0,03.36,5 = 1,095{\text{ gam}} \to {\text{C}}{{\text{\% }}_{HCl{\text{ dư}}}} = \frac{{1,095}}{{89,75}} = 1,22\% \)