Đáp án:

\(\begin{array}{l}
\% {m_{Al}} = 21,51\% \\
\% {m_{BaC{O_3}}} = 78,49\% \\
a = 17,22g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
BaC{O_3} + 2HCl \to BaC{l_2} + C{O_2} + {H_2}O\\
{n_{HCl}} = 0,06 \times 2 = 0,12\,mol\\
{n_{hh}} = \dfrac{{0,896}}{{22,4}} = 0,04\,mol\\
hh:Al(a\,mol),BaC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
27a + 197b = 2,51\\
1,5a + b = 0,04
\end{array} \right.\\
 \Rightarrow a = 0,02;b = 0,01\\
\% {m_{Al}} = \dfrac{{0,02 \times 27}}{{2,51}} \times 100\%  = 21,51\% \\
\% {m_{BaC{O_3}}} = 100 - 21,51 = 78,49\% \\
{n_{HCl}} = 0,12 - 0,04 \times 2 = 0,04\,mol\\
AgN{O_3} + HCl \to AgCl + HN{O_3}\\
AlC{l_3} + 3AgN{O_3} \to 3AgCl + Al{(N{O_3})_3}\\
BaC{l_2} + 2AgN{O_3} \to 2AgCl + Ba{(N{O_3})_2}\\
{n_{AgCl}} = 0,04 + 0,02 \times 3 + 0,01 \times 2 = 0,12\,mol\\
a = {m_{AgCl}} = 0,12 \times 143,5 = 17,22g
\end{array}\)