Đáp án:

a, \(Zn\)

b,

\({m_{HCl}}{\rm{dd = }}\dfrac{{18,25 \times 100}}{{18,25}} = 100g\)

\(C{M_{HCl}} = \dfrac{{0,5}}{{0,083}} = 6M\)

c, \(C\% ZnC{l_2} = \dfrac{{34}}{{115,75}} \times 100\%  = 29,37\% \)

Giải thích các bước giải:

Gọi kim loại là R

\(\begin{array}{l}
R + 2HCl \to RC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,25mol\\
 \to {n_R} = {n_{{H_2}}} = 0,25mol\\
 \to {M_R} = 65 \to Zn\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,5mol\\
 \to {m_{HCl}} = 18,25g\\
 \to {m_{HCl}}{\rm{dd = }}\dfrac{{18,25 \times 100}}{{18,25}} = 100g\\
 \to {V_{HCl}}dd = \dfrac{{100}}{{1,2}} = 83,33ml\\
 \to C{M_{HCl}} = \dfrac{{0,5}}{{0,083}} = 6M\\
{m_{dd}} = {m_{KL}} + {m_{HCl}}dd - {m_{{H_2}}} = 16,25 + 100 - 0,25 \times 2 = 115,75g\\
{n_{ZnC{l_2}}} = {n_{{H_2}}} = 0,25mol\\
 \to {m_{ZnC{l_2}}} = 34g\\
 \to C\% ZnC{l_2} = \dfrac{{34}}{{115,75}} \times 100\%  = 29,37\% 
\end{array}\)