$n_{CaO}=$ $\dfrac{5,6}{56}=0,1 (mol)$
$\text{ PTHH }$
$\text{CaO + 2HCl → }$ $H_{2}O+$ $CaCl_{2}$
$\text{ theo PTHH : }$
$n_{HCl}=2.$ $n_{CaO}=2.0,1=0,2(mol)$
$=>m_{HCl}=n.M=0,2.36,5=7,3gam$
$\text{Ta có C% = }$ $m_{chất tan }:$ $m_{dung dịch}$
$m_{dung dịch}=$ $m_{chất tan }: C $ $\text{ % }=7,3 : 14,6$ $\text{% = 50 g}$