Sửa đề : 7.3%
\(m_{HCl}=100\cdot7.3\%=7.3\left(g\right)\)
\(n_{HCl}=\dfrac{7.3}{36.5}=0.2\left(mol\right)\)
\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+H_2O\)
\(0.1..............0.2\)
\(m_{Ca\left(OH\right)_2}=0.1\cdot74=7.4\left(g\right)\)
\(m_{dd}=m_{Ca\left(OH\right)_2}+m_{dd_{HCl}}=7.4+100=107.4\left(g\right)\)