Đáp án:
\(C{\% _{AlC{l_3}}} = 12,7\% \)
Giải thích các bước giải:
\(\begin{array}{l}
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{A{l_2}{O_3}}} = \dfrac{{20,4}}{{102}} = 0,2mol\\
{m_{HCl}} = \dfrac{{200 \times 10,95}}{{100}} = 21,9g\\
{n_{HCl}} = \dfrac{{21,9}}{{36,5}} = 0,6mol\\
\dfrac{{0,2}}{1} > \dfrac{{0,6}}{6} \Rightarrow A{l_2}{O_3}\text{ dư}\\
{n_{A{l_2}{O_3}pu}} = \dfrac{{{n_{HCl}}}}{6} = 0,1mol\\
{m_{A{l_2}{O_3}pu}} = 0,1 \times 102 = 10,2g\\
{m_{ddspu}} = 200 + 10,2 = 210,2g\\
{n_{AlC{l_3}}} = \dfrac{{{n_{HCl}}}}{3} = 0,2mol\\
{m_{AlC{l_3}}} = 0,2 \times 133,5 = 26,7g\\
C{\% _{AlC{l_3}}} = \dfrac{{26,7}}{{210,2}} \times 100\% = 12,7\%
\end{array}\)
