Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 5,6l\\
b)\\
{m_{HCl}} = 18,25g\\
c)\\
{A_{Zn}} = 1,5 \times {10^{23}}\\
d)\\
R:Cu
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{{16,25}}{{65}} = 0,25\,mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,25\,mol\\
{V_{{H_2}}} = 0,25 \times 22,4 = 5,6l\\
b)\\
{n_{HCl}} = 2{n_{Zn}} = 0,5\,mol\\
{m_{HCl}} = 0,5 \times 36,5 = 18,25g\\
c)\\
{A_{Zn}} = 0,25 \times 6 \times {10^{23}} = 1,5 \times {10^{23}}\\
d)\\
RO + {H_2} \to R + {H_2}O\\
{n_{RO}} = {n_{{H_2}}} = 0,25\,mol\\
{M_{RO}} = \dfrac{{20}}{{0,25}} = 80\,g/mol \Rightarrow {M_R} = 80 - 16 = 64\,g/mol\\
\Rightarrow R:Cu
\end{array}\)