Sục 7,84 lít $C{{O}_{2}}$ (đktc) vào 250 ml dung dịch $Ca{{\left(...

${{n}_{O{{H}^{-}}}}=2{{n}_{Ca{{\left( OH \right)}_{2}}}}=2.0,25=0,5 mol$
Ta có: $\dfrac{{{n}_{O{{H}^{-}}}}}{{{n}_{C{{O}_{2}}}}}=\dfrac{0,5}{0,35}\approx 1,4$
Nên sau phản ứng tạo 2 muối $Ca{{\left( HC{{O}_{3}} \right)}_{2}}, CaC{{O}_{3}}$
$\begin{aligned}
& C{{O}_{2}}+O{{H}^{-}}\xrightarrow{{}}HCO_{3}^{-} \\
& x x x \\
\end{aligned}$
$\begin{aligned}
& C{{O}_{2}}+2O{{H}^{-}}\xrightarrow{{}}CO_{3}^{2-}+{{H}_{2}}O \\
& y 2y y \\
\end{aligned}$
Ta có: $\left\{ \begin{aligned}
& HCO_{3}^{-}:x mol \\
& CO_{3}^{2-}:y mol \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x+2y=0,5\left( {{n}_{O{{H}^{-}}}} \right) \\
& x+y=0,35 \left( {{n}_{C{{O}_{2}}}} \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=0,2 \\
& y=0,15 \\
\end{aligned} \right.$
$\begin{aligned}
& C{{a}^{2+}}+CO_{3}^{2-}\xrightarrow{{}}CaC{{O}_{3}} \\
& 0,15 0,15 \\
\end{aligned}$
$\Rightarrow {{m}_{CaC{{O}_{3}}}}=15 gam$