\(n_{CO_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(n_{KOH}=\dfrac{400\cdot5.6\%}{56}=0.4\left(mol\right)\)
\(n_{K_2CO_3}=a\left(mol\right),n_{KHCO_3}=b\left(mol\right)\)
\(2KOH+CO_2\rightarrow K_2CO_3+H_2O\)
\(KOH+CO_2\rightarrow KHCO_3\)
\(\left\{{}\begin{matrix}2a+b=0.4\\a+b=0.3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.2\end{matrix}\right.\)
\(K_2CO_3+BaCl_2\rightarrow BaCO_3+2KCl\)
\(0.1...............................0.1\)
\(m_{BaCO_3}=0.1\cdot197=19.7\left(g\right)\)
Ta có : \(n_{CO2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{KOH}=\dfrac{400.5,6}{56.100}=0,4\left(mol\right)\)
\(\dfrac{n_{KOH}}{n_{CO2}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\Rightarrow\)Tạo 2 muối
\(CO2+2KOH\rightarrow K2CO3+H2O\)
x---------->2x-------->x(mol)
\(CO2+KOH\rightarrow KHCO3\)
y-------->y------------->(mol)
Theo bài ta có HPT\(\left\{{}\begin{matrix}x+y=0,3\\2x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(K2CO3+BaCl2\rightarrow BaCO3+2KCl\)
0,1--------------------------->0,1(mol)
\(\Rightarrow m=m_{BaCO3}=0,1.197=19,7\left(g\right)\)
Chúc bạn học tốt ^.^
