Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{2,7}}{{27}} = 0,1\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1\,mol\\
{m_{AlC{l_3}}} = n \times M = 0,1 \times 133,5 = 13,35g\\
c)\\
{n_{{H_2}}} = \dfrac{{0,1 \times 3}}{2} = 0,15\,mol\\
{V_{{H_2}}} = n \times 22,4 = 0,15 \times 22,4 = 3,36l\\
d)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,3\,mol\\
{m_{HCl}} = n \times M = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{ddHCl}}}} = \dfrac{{m \times 100}}{{{C_\% }}} = \dfrac{{10,95 \times 100}}{{3,65}} = 300g\\
{m_{{\rm{dd}}spu}} = {m_{Al}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 2,7 + 300 - 0,15 \times 2 = 302,4g
\end{array}\)
