Đáp án:
\(\begin{array}{l}
b)\\
{V_{{O_2}}} = 1,68l\\
c)\\
{m_{{\rm{dd}}HCl}} = 75g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
b)\\
{n_{Al}} = \dfrac{{2,7}}{{27}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{3}{4}{n_{Al}} = 0,075mol\\
{V_{{O_2}}} = 0,075 \times 22,4 = 1,68l\\
c)\\
{n_{A{l_2}{O_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,05mol\\
{n_{HCl}} = 6{n_{A{l_2}{O_3}}} = 0,3mol\\
{m_{HCl}} = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{10,95 \times 100}}{{14,6}} = 75g
\end{array}\)