$\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{4}{1}\Rightarrow {{m}_{2}}=\frac{1}{4}{{m}_{1}}=\frac{1}{4}\times \frac{8}{13}m=\frac{2}{13}m$$\Rightarrow $ Khối lượng kết tủa bị hoàn tan: $\frac{11}{13}m$


+ Khi cho V (l) CO$_{2}$
$C{{O}_{2}}+Ca{{(OH)}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O$
$n\downarrow ={{n}_{C{{O}_{2}}}}=\frac{{{m}_{1}}}{100}=\frac{2}{325}m=\frac{V}{22,4}$ (*)
+ Khi cho V + 2,688 (l) CO$_{2}$
$C{{O}_{2}}+Ca{{(OH)}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O$ (1)
${{n}_{C{{O}_{2}}(1)}}={{n}_{CaC{{O}_{3}}(1)}}=\frac{m}{100}$
$C{{O}_{2}}+CaC{{O}_{3}}+{{H}_{2}}O\to Ca{{(HC{{O}_{3}})}_{2}}$ (2)
${{n}_{C{{O}_{2}}(2)}}={{n}_{CaC{{O}_{3}}(2)}}=\frac{\frac{11}{13}m}{100}=\frac{11}{1300}m$
$\sum{{{n}_{C{{O}_{2}}}}=\frac{m}{100}+\frac{11m}{1300}=\frac{6}{325}m=\frac{V+2,688}{22,4}}$ (**)
Từ (*) và (**) $\Rightarrow $ V = 1,344 (l), m = 9,75 (g)
+ Khi cho V + V$_{1}$ (l) CO$_{2}$
$C{{O}_{2}}+Ca{{(OH)}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O$
$n\downarrow \max =\frac{9,75}{100}=0,0975(mol)={{n}_{C{{O}_{2}}}}$
${{V}_{C{{O}_{2}}}}=2,184(l)\Rightarrow {{V}_{1}}=2,184-1,344=0,84(l)$