$\text{PTHH: Zn + 2HCl -> $ZnCl_2$ + $H_2$}$

$\text{b,}$

$\text{nZn = $\frac{13}{65}$ = 0,2(mol)}$

$\text{$nH_2$  = nZn = 0,2(mol) }$

$\text{=> $V_{H_2}$ = 0,2 * 22,4 = 4,48(lít)}$

$\text{c,}$

$\text{-Muối là $ZnCl_2$}$

$\text{=> $n_{ZnCl_2}$ = nZn= 0,2(mol) }$

$\text{=> m = 0,2 * 136 = 27,2(g)}$

$\text{c, CuO + $H_2$ $\xrightarrow[]{t^o}$ Cu + $H_2O$}$

Theo phương trình => $nCu$ = $nH_2$ = $0,2$ (mol)

=> $mCu$ = $0,2$ * $64$ = $12,8$ (g)

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{13}}{{65}} = 0,2\,mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,2\,mol\\
{m_{ZnC{l_2}}} = 0,2 \times 136 = 27,2g\\
d)\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{n_{Cu}} = {n_{{H_2}}} = 0,2\,mol\\
{m_{Cu}} = 0,2 \times 64 = 12,8g
\end{array}\)