Đáp án:
a)
\(\begin{array}{l}
\% {m_{Cu}} = 65,31\% \\
\% {m_{Al}} = 18,37\% \\
\% {m_{Mg}} = 16,32\%
\end{array}\)
b) 9,125%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{m_{Cu}} = {m_{cr}} = 19,2g\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
27a + 24b = 29,4 - 19,2\\
1,5a + b = 0,5
\end{array} \right.\\
\Rightarrow a = b = 0,2\,mol\\
\% {m_{Cu}} = \dfrac{{19,2}}{{29,4}} \times 100\% = 65,31\% \\
\% {m_{Al}} = \dfrac{{0,2 \times 27}}{{29,4}} \times 100\% = 18,37\% \\
\% {m_{Mg}} = 100 - 65,31 - 18,37 = 16,32\% \\
b)\\
AlC{l_3} + 3AgN{O_3} \to 3AgCl + Al{(N{O_3})_3}\\
MgC{l_2} + 2AgN{O_3} \to 2AgCl + Mg{(N{O_3})_2}\\
HCl + AgN{O_3} \to AgCl + HN{O_3}\\
{n_{AgCl}} = \dfrac{{200,9}}{{143,5}} = 1,4\,mol\\
{n_{HCl}} \text{ dư }= 1,4 - 0,2 \times 3 - 0,2 \times 2 = 0,4\,mol\\
{n_{HCl}} \text{ phản ứng }= 2{n_{{H_2}}} = 1\,mol\\
{n_{HCl}} \text{ ban đầu }= 0,4 + 1 = 1,4\,mol\\
{m_{{\rm{dd}}HCl}} = 500 \times 1,12 = 560g\\
{C_\% }HCl = \dfrac{{1,4 \times 36,5}}{{560}} \times 100\% = 9,125\%
\end{array}\)
