Đáp án:
\(\begin{array}{l}
\% Cu = 32,53\% \\
\% Fe = 42,69\% \\
\% Zn = 24,78\% \\
m = 96,15g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{m_{Cu}} = 12,8g\\
{n_{{H_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45mol\\
hh:Fe(a\,mol),Zn(b\,mol)\\
a + b = 0,45(1)\\
56a + 65b = 26,55(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,3mol;b = 0,15mol\\
{m_{Fe}} = 0,3 \times 56 = 16,8g\\
{m_{Zn}} = 0,15 \times 65 = 9,75g\\
\% Cu = \dfrac{{12,8}}{{39,35}} \times 100\% = 32,53\% \\
\% Fe = \dfrac{{16,8}}{{39,35}} \times 100\% = 42,69\% \\
\% Zn = \dfrac{{9,75}}{{39,35}} \times 100\% = 24,78\% \\
Cu + C{l_2} \to CuC{l_2}\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
Zn + C{l_2} \to ZnC{l_2}\\
{n_{Cu}} = \dfrac{{12,8}}{{64}} = 0,2mol\\
{n_{CuC{l_2}}} = {n_{Cu}} = 0,2mol\\
{m_{CuC{l_2}}} = 0,2 \times 135 = 27g\\
{n_{FeC{l_3}}} = {n_{Fe}} = 0,3mol\\
{m_{FeC{l_3}}} = 0,3 \times 162,5 = 48,75g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,15mol\\
{m_{ZnC{l_2}}} = 0,15 \times 136 = 20,4g\\
m = 27 + 48,75 + 20,4 = 96,15g
\end{array}\)
