$n_{CaO} = \dfrac{5,6}{56} =0,1(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
Theo PTHH :
$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
\(n_{CaO}\) = 0,1 mol
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
0,1 mol → 0,2 mol
\(\Rightarrow m_{HCl}\) = 0,2.36,5 = 7,3 gam
\(\Rightarrow\) Khối lượng dd HCl đã dùng là: \(m_{HCl}\)= \(\dfrac{7,3\times100\%}{14,6\%}\)=50(g)