a. $Zn+2HCl\to ZnCl_2+H_2$
b.
$n_{Zn}=\dfrac{13}{65}=0,2(mol)$
$\to n_{HCl}=2n_{Zn}=0,4(mol)$
$\to m_{HCl}=36,5.0,4=14,6(g)$
c. Cách 1:
$n_{Zn}=n_{ZnCl_2}=0,2(mol)$
$\to m_{ZnCl_2}=0,2.136=27,2(g)$
Cách 2:
BTKL: $m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}$
$\to m_{ZnCl_2}=13+14,6-0,4=27,2(g)$