Đáp án:
\(\begin{array}{l}
a)\\
M:\text{Nhôm}(Al)\\
b)\\
{m_{ddHCl}} = 284,7g\\
c)\\
C{\% _{HCl}} = 0,45\% \\
C{\% _{AlC{l_3}}} = 13,72\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2M + 6HCl \to 2MC{l_3} + 3{H_2}(1)\\
{n_{{H_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45mol\\
{n_M} = \dfrac{2}{3}{n_{{H_2}}} = 0,3mol\\
{M_M} = \dfrac{{8,1}}{{0,3}} = 27dvC\\
\Rightarrow M:\text{Nhôm}(Al)\\
b)\\
{n_{HCl(1)}} = 2{n_{{H_2}}} = 0,9mol\\
{m_{HCl(1)}} = 0,9 \times 36,5 = 32,85g\\
{m_{HC{l_d}}} = \dfrac{{32,85 \times 4}}{{100}} = 1,314g\\
{m_{HCl}} = 32,85 + 1,314 = 34,164g\\
{m_{ddHCl}} = \dfrac{{34,164 \times 100}}{{12}} = 284,7g\\
c)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,3mol\\
{m_{AlC{l_3}}} = 0,3 \times 133,5 = 40,05g\\
{m_{ddspu}} = 8,1 + 284,7 - 0,45 \times 2 = 291,9g\\
C{\% _{HCl}} = \dfrac{{1,314}}{{291,9}} \times 100\% = 0,45\% \\
C{\% _{AlC{l_3}}} = \dfrac{{40,05}}{{291,9}} \times 100\% = 13,72\%
\end{array}\)
