Sục 7,84 lít CO2 (đktc) vào 250 ml dung dịch Ca(OH)2 1M. Khối lượng kết tủa thu

Đáp án B

$n_{O{H}^{-}}=2n_{Ca{\left(OH\right)}_2}=2.0,25=0,5mol$

Ta có: $\frac{n_{O{H}^{-}}}{n_{C{O}_2}}=\frac{0,5}{0,35}\approx 1,4$

Nên sau phản ứng tạo 2 muối $Ca{\left(HC{O}_3\right)}_2,CaC{O}_3$

$$\begin{gathered} \begin{array}{l}C{O}_2+O{H}^{-}\stackrel{}{\to }HC{O}_3^{-}\\ x x x\end{array} \\ \begin{array}{l}C{O}_2+2O{H}^{-}\stackrel{}{\to }C{O}_3^{2-}+H_{2}O\\ y2yy\end{array} \end{gathered}$$

Ta có: $\left\{\begin{array}{l}HC{O}_3^{-}:xmol\\ C{O}_3^{2-}:ymol\end{array}\right.\Rightarrow \left\{\begin{array}{l}x+2y=0,5\left(n_{O{H}^{-}}\right)\\ x+y=0,35\left(n_{C{O}_2}\right)\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,2\\ y=0,15\end{array}\right.$

$$\begin{gathered} \begin{array}{l}C{a}^{2+}+C{O}_3^{2-}\stackrel{}{\to }CaC{O}_3\\ 0,15 0,15\end{array} \\ \Rightarrow m_{CaC{O}_3}=15gam \end{gathered}$$