Phản ứng xảy ra:
\(2{\text{A}}l + 6HCl\xrightarrow{{}}2{\text{A}}lC{l_3} + 3{H_2}\)
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Gọi số mol \(Al;Zn\) lần lượt là \(x;y\)
\( \to 27x+65y=18,4 \text{ gam}\)
Ta có:
\({n_{{H_2}}} = \frac{3}{2}x + y = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol}}\)
Giải được:
\(x=y=0,2\)
\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{5,4}}{{18,4}}.100\% = 29,35\% \to \% {m_{Zn}} = 70,65\% \)
Mặt khác:
\({n_{AlC{l_3}}} = {n_{Al}} = 0,2{\text{ mol;}}{{\text{n}}_{ZnC{l_2}}} = {n_{Zn}} = 0,2{\text{ mol}}\)
\( \to {m_{muối}} = {m_{AlC{l_3}}} + {m_{ZnC{l_2}}} = 0,2.133,5 + 0,2.136 = 53,9{\text{ gam}}\)