Đáp án:
\(Fe:{\text{ x mol; Zn: y mol}}\)
\(Fe + 2HCl \to FeCl_2^{} + H_2^{}\)
\(\;{\text{x 2x x}}\)
\(Zn + 2HCl \to ZnCl_2^{} + H_2^{}\)
\(\;{\text{y 2y y}}\)
\(56x + 65y = 8,85\)
\(x + y = 0,15\)
\( \to x = 0,1\;{\text{mol; y = 0}}{\text{,05 mol}}\)
\({\text{\% m}}_{Fe}^{} = \frac{{0,1.56}}{{8,85}}.100 = 63,27\% \)
\(\% m_{Zn}^{} = 100 - 63,27 = 36,73\% \)
\(b)\)
\(n_{HCl}^{} = 0,1.2 + 0,05.2 = 0,3\;{\text{mol}}\)
\({\text{m}}_{HCl}^{} = 0,3.36,5 = 10,95{\text{ gam}}\)
\(m_{dd_{HCl}^{}}^{} = \frac{{10,95}}{{10}}.100 = 109,5{\text{ gam}}\)