$n_{OH^-}=2n_{Ca(OH)_2}=2b\ (mol)$
$T=\dfrac{n_{OH^-}}{n_{CO_2}}=\dfrac{2b}{a}$
Do $b<a<2b\to 1<T<2$
$\to$ Phản ứng tạo hai muối
Ta có: $\left\{\begin{matrix}n_{CO_3^{2-}}=n_{OH^-}-n_{CO_2}=2b-a\ (mol)\\n_{HCO_3^-}=2n_{CO_2}-n_{OH^-}=2a-2b\ (mol)\end{matrix}\right.$
$\to \left\{\begin{matrix}m_{\downarrow}=m_{CaCO_3}=100(2b-a)\ gam\\m_{muối}=m_{Ca(HCO_3)_2}=162(a-b)\ gam\end{matrix}\right.$