Đáp án:
\(\begin{array}{l}
b)\\
{m_{A{l_2}{O_3}}} = 8,5g\\
{m_{AlC{l_3}}} = 22,25g\\
c)\\
{m_{{H_2}O}} = 4,5g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
b)\\
{n_{HCl}} = \dfrac{{18,25}}{{36,5}} = 0,5\,mol\\
{n_{A{l_2}{O_3}}} = 0,5 \times \dfrac{1}{6} = \dfrac{1}{{12}}\,mol\\
{m_{A{l_2}{O_3}}} = \dfrac{1}{{12}} \times 102 = 8,5g\\
{n_{AlC{l_3}}} = 0,5 \times \dfrac{2}{6} = \dfrac{1}{6}\,mol\\
{m_{AlC{l_3}}} = \dfrac{1}{6} \times 133,5 = 22,25g\\
c)\\
{m_{{H_2}O}} = {m_{A{l_2}{O_3}}} + {m_{HCl}} - {m_{AlC{l_3}}}\\
{m_{{H_2}O}} = 8,5 + 18,25 - 22,25 = 4,5g
\end{array}\)