Chọn A.

Sửa đề : Ca(OH)₂ 0.01 M

\(n_{CO_2}=\dfrac{0.2688}{22.4}=0.012\left(mol\right)\)

\(n_{OH^-}=0.2\cdot0.1+0.2\cdot0.01\cdot2=0.024\left(mol\right)\)

\(\dfrac{n_{OH^-}}{n_{CO_2}}=\dfrac{0.024}{0.012}=2\)

\(\Rightarrow\text{Tạo ra muối trung hòa}\)

\(2OH^-+CO_2\rightarrow CO_3^{2-}+H_2O\)

\(0.024......0.012.......0.012\)

\(m_M=m_{Ca^{2+}}+m_{Na^+}+m_{CO_3^{2-}}=0.002\cdot40+0.02\cdot23+0.012\cdot60=1.26\left(g\right)\)

\(n_{CO_2}=\dfrac{2.688}{22.4}=0.12\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0.1\cdot1=0.1\left(mol\right)\)

\(T=\dfrac{0.12}{0.1}=1.2\)

=> Tạo 2 muối

\(n_{CaCO_3}=a\left(mol\right),n_{Ca\left(HCO_3\right)_2}=b\left(mol\right)\)

Ta có : 

\(a+b=0.1\)

\(a+2b=0.12\)

\(\Rightarrow\left\{{}\begin{matrix}a=0.08\\b=0.02\end{matrix}\right.\)

\(m_{Ca\left(HCO_3\right)_2}=0.02\cdot162=3.24\left(g\right)\)

\(n_{CO_2}=\dfrac{5.04}{22.4}=0.225\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0.25\cdot0.5=0.125\left(mol\right)\)

\(T=\dfrac{0.225}{0.125}=1.8\)

=> Tạo ra 2 muối

\(n_{BaCO_3}=a\left(mol\right),n_{Ba\left(HCO_3\right)_2}=b\left(mol\right)\)

Ta có : 

\(a+b=0.125\)

\(a+2b=0.225\)

\(\Rightarrow\left\{{}\begin{matrix}a=0.025\\b=0.1\end{matrix}\right.\)

\(m_{Muối}=0.025\cdot197+0.1\cdot259=30.825\left(g\right)\)

Đáp án : A

Ta có : n CO2 = 0,2688 / 22,4 = 0,012 mol

n _NaOH = 0,2.0,1 = 0,02 mol     => n Na⁺  = 0,02 mol ; nOH^- = 0,024 mol

n Ca(OH)₂ = 0,2.0,01 = 0,002 mol     => n Ca²⁺ = 0,002 mol

ta thấy  k = nOH^- / nCO₂  =  0,024 / 0,012 = 2 => tạo ion CO₃^2-

 CO₂ + 2OH ^- → CO₃^2-  + H2O

0,012               0,012

=> m _muối = m _cation + m _anion =  m Ca²⁺ + mNa⁺ + m CO₃^2-

                   =  0,002.40 + 0,02.23 + 0,012.60 = 1,26 gam

Sửa đề : 0.01 (M) 

\(n_{CO_2}=\dfrac{0.2688}{22.4}=0.012\left(mol\right)\)

\(n_{OH^-}=0.2\cdot0.1+0.2\cdot0.01\cdot2=0.024\left(mol\right)\)

\(\dfrac{n_{OH^-}}{n_{CO_2}}=\dfrac{0.024}{0.012}=2\)

=> Phản ứng tạo ra muối trung hòa

\(n_{CO_2}=n_{H_2O}=0.012\left(mol\right)\)

\(\text{Bảo toàn khối lượng : }\)

\(m_{Bazo}+m_{CO_2}=m_M+m_{H_2O}\)

\(\Rightarrow m_M=0.02\cdot40+0.002\cdot74+0.012\cdot44-0.012\cdot18=1.26\left(g\right)\)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Xét \(T=\dfrac{0,5.n_{Ca\left(OH\right)_2}}{n_{CO_2}}\)

Do pư chỉ tạo muối CaCO₃

=> \(T\ge2\)

=> \(\dfrac{0,5.n_{Ca\left(OH\right)_2}}{0,1}\ge2\Rightarrow n_{Ca\left(OH\right)_2}\ge0,4\left(mol\right)\)

=> Chỉ có A thỏa mãn

A

a