Đáp án:
a) \(\% {m_{Al}} = 40\% ;\% {m_{ZnO}} = 60\% \)
b) \({{\text{m}}_{dd{\text{ HCl}}}} = 200{\text{ gam}}\)
c) \(C{\% _{AlC{l_3}}} = 23,65\% ;{\text{ C}}{{\text{\% }}_{ZnC{l_2}}} = 12,05\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(ZnO + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol = }}\frac{3}{2}{n_{Al}} \to {n_{Al}} = 0,4{\text{ mol}} \to {{\text{m}}_{Al}} = 0,4.27 = 10,8{\text{ gam}} \to {{\text{m}}_{ZnO}} = 27 - 10,8 = 16,2{\text{ gam}}\)
\(\to \% {m_{Al}} = \frac{{10,8}}{{27}} = 40\% \to \% {m_{ZnO}} = 60\% \)
\(\to {n_{ZnO}} = \frac{{16,2}}{{65 + 16}} = 0,2{\text{ mol}}\)
Ta có:
\({n_{HCl}} = 3{n_{Al}} + 2{n_{ZnO}} = 0.4.3 + 0.2.2 = 1,6{\text{ mol}} \to {{\text{m}}_{HCl}} = 1,6.36,5 = 58,4{\text{ gam}} \to {{\text{m}}_{dd{\text{ HCl}}}} = \frac{{58,4}}{{29,2\% }} = 200{\text{ gam}}\)
BTKL: \({m_{dd{\text{ sau phản ứng}}}} = {m_{hh}} + {m_{HCl}} - {m_{{H_2}}} = 27 + 200 - 0,6.2 = 225,8{\text{ gam}}\)
\({n_{AlC{l_3}}} = {n_{Al}} = 0,4{\text{ mol; }}{{\text{n}}_{ZnC{l_2}}} = {n_{ZnO}} = 0,2{\text{ mol}}\)
\( \to {m_{AlC{l_3}}} = 0,4.(27 + 35,5.3) = 53,4{\text{ gam; }}{{\text{m}}_{ZnC{l_2}}} = 0,2.(65 + 35,5.2) = 27,2{\text{ gam}}\)
\( \to C{\% _{AlC{l_3}}} = \frac{{53,4}}{{225,8}} = 23,65\% ;{\text{ C}}{{\text{\% }}_{ZnC{l_2}}} = \frac{{27,2}}{{225,8}} = 12,05\% \)
