Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 6,72l\\
b)\\
{m_{{\rm{dd}}HCl}} = 150g\\
c)\\
{C_\% }AlC{l_3} = 17,25\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
{n_{{H_2}}} = 0,2 \times \dfrac{3}{2} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,6 \times 36,5}}{{14,6\% }} = 150g\\
c)\\
{m_{{\rm{dd}}spu}} = 5,4 + 150 - 0,3 \times 2 = 154,8g\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,2\,mol\\
{C_\% }AlC{l_3} = \dfrac{{0,2 \times 133,5}}{{154,8}} \times 100\% = 17,25\%
\end{array}\)
