Đáp án:
\(\begin{array}{l}
a)\\
\% Zn = 81,04\% \\
\% Al = 18,96\% \\
b)\\
{C_{{M_{ZnC{l_2}}}}} = 0,38M\\
{C_{{M_{AlC{l_3}}}}} = 0,214M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{HCl}} = 0,5 \times 1,4 = 0,7mol\\
hh:Zn(a\,mol),Al(b\,mol)\\
136a + 133,5b = 40,05(1)\\
2a + 3b = 0,7(2)\\
\text{Từ (1) và (2)} \Rightarrow a = 0,19;b = 0,107\\
{m_{Zn}} = 0,19 \times 65 = 12,35g\\
{m_{Al}} = 0,107 \times 27 = 2,889g\\
\% Zn = \dfrac{{12,35}}{{12,35 + 2,889}} \times 100\% = 81,04\% \\
\% Al = 100 - 81,04 = 18,96\% \\
b)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,19mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,107mol\\
{C_{{M_{ZnC{l_2}}}}} = \dfrac{{0,19}}{{0,5}} = 0,38M\\
{C_{{M_{AlC{l_3}}}}} = \dfrac{{0,107}}{{0,5}} = 0,214M
\end{array}\)
