Đáp án:

 1) 3,49g

2) 11,7g

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
AC{O_3} + 2HCl \to AC{l_2} + C{O_2} + {H_2}O\\
BC{O_3} + 2HCl \to BC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{0,672}}{{22,4}} = 0,03\,mol\\
{n_{HCl}} = 2{n_{C{O_2}}} = 0,03 \times 2 = 0,06\,mol\\
{n_{{H_2}O}} = {n_{C{O_2}}} = 0,03\,mol\\
{m_X} + {m_{HCl}} = {m_Y} + {m_{{H_2}O}}\\
 \Rightarrow {m_Y} = 1,84 + 0,06 \times 36,5 - 0,03 \times 18 = 3,49g\\
2)\\
{n_{N{a_2}O}} = \dfrac{{15,5}}{{62}} = 0,25\,mol\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{NaOH}} = 0,25 \times 2 = 0,5\,mol\\
{n_{AlC{l_3}}} = 0,15 \times 1 = 0,15\,mol\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
\dfrac{{0,15}}{1} < \dfrac{{0,5}}{3} \Rightarrow \text{ NaOH dư}\\
{n_{Al{{(OH)}_3}}} = {n_{AlC{l_3}}} = 0,15\,mol\\
{m_{Al{{(OH)}_3}}} = 0,15 \times 78 = 11,7g
\end{array}\)