Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{13}}{{65}} = 0,2mol\\
{n_{HCl}} = 2{n_{Zn}} = 0,4mol\\
{m_{HCl}} = n \times M = 0,4 \times 36,5 = 14,6g\\
c)\\
{n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
{V_{Zn}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
2)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{44,8}}{{22,4}} = 2mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 4mol\\
{m_{HCl}} = n \times M = 4 \times 36,5 = 146g\\
c)\\
{n_{MgC{l_2}}} = {n_{H{  _2}}} = 2mol\\
{m_{MgC{l_2}}} = n \times M = 2 \times 95 = 190g\\
3)\\
CTHH:{H_x}{S_y}\\
\% H = \dfrac{{x{M_H}}}{M} \times 100\%  = 5,88\% \\
 \Rightarrow x = 34 \times 0,0588 = 2\\
\% S = \dfrac{{y{M_S}}}{M} \times 100\%  = 94,12\% \\
 \Rightarrow y = \dfrac{{34 \times 0,9412}}{{32}} = 1\\
 \Rightarrow CTHH:{H_2}S
\end{array}\)

Bài giải :

Câu 1 :

a. −nZn=1365=0,2(mol)-nZn=1365=0,2(mol)

Zn+2HCl→ZnCl2+H2↑Zn+2HCl→ZnCl2+H2↑

 0,2   →   0,4           0,2          0,2             (mol)