Đáp án:
\(\begin{array}{l}
a)\\
{m_{Zn}} = 13g\\
b)\\
{m_{Al}} = 3,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{V_{{H_2}}} = 4,48d{m^3} = 4,48l\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,2mol\\
{m_{Zn}} = 0,2 \times 65 = 13g\\
b)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = \dfrac{2}{{15}}mol\\
{m_{Al}} = \dfrac{2}{{15}} \times 27 = 3,6g
\end{array}\)