$PTHH:CO_2+Ca(OH)_2\to CaCO_3+H_2O\\ n_{CO_2}=\dfrac{0,4958}{24,79}=0,02(mol)\\ n_{Ca(OH)_2}=\dfrac{44,4.5\%}{74.100\%}=0,03(mol)\\ \text{Lập tỉ lệ: }\dfrac{n_{CO_2}}{1}<\dfrac{n_{Ca(OH)_2}}{1}\Rightarrow Ca(OH)_2\text{ dư}\\ a,n_{CaCO_3}=n_{H_2O}=n_{CO_2}=0,02(mol)\\ \Rightarrow m_{CaCO_3}=0,02.100=2(g);m_{H_2O}=0,02.18=0,36(g)\\ b,m_{CO_2}=0,02.44=0,88(g)\\ n_{Ca(OH)_2(dư)}=0,03-0,02=0,01(mol)\\ \Rightarrow m_{Ca(OH)_2(dư)}=0,01.74=0,74(g)\\ \Rightarrow \begin{cases} C\%_{Ca(OH)_2(dư)}=\dfrac{0,74}{0,88+44,4}.100\%\approx 1,63\%\\ C\%_{CaCO_3}=\dfrac{2}{0,88+44,4}.100\%=4,42\% \end{cases}\\ c,PTHH:Ca(OH)_2+2HCl\to CaCl_2+2H_2O\\ \Rightarrow n_{HCl}=2n_{Ca(OH)_2(bđ)}=2.0,03=0,06(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,06.36,5}{20\%}=10,95(g)$