a) Gọi số mol CO2 và SO2 là a,b (mol)
Ta có: \(\overline{M_X}=\frac{44a+64b}{a+b}=26.2\) => a = 1,5b
\(=>\left\{{}\begin{matrix}\%n_{CO_2}=\frac{a}{a+b}.100\%=\frac{1,5b}{1,5b+b}.100\%=60\%\\\%n_{SO_2}=\frac{b}{a+b}.100\%=\frac{b}{1,5b+b}.100\%=40\%\end{matrix}\right.\)
b) \(n_X=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Mà a = 1,5b
=> \(\left\{{}\begin{matrix}a=0,3\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)
__________________ 0,3 --------> 0,3_______(mol)
\(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3+H_2O\)
____________0,2 ------> 0,2_________(mol)
=> \(\left\{{}\begin{matrix}m_{BaCO_3}=0,3.197=59,1\left(g\right)\\m_{BaSO_3}=0,2.217=43,4\left(g\right)\end{matrix}\right.\)
=> \(m_{KT}=59,1+43,4=102,5\left(g\right)\)
