a)\(CaCO3-->CaO+CO2\)
x-----------------------------------x(mol)
\(BaCO3-->BaO+CO2\)
y----------------------------y(mol)
b)Gọi n CaCO3= x, n BaCO3 =y
\(n_{CO2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}100x+197y=7,94\\x+y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,02\end{matrix}\right.\)
\(\%m_{CaCO3}=\frac{100.0,4}{7,94}.100\%=50,38\%\)
%\(m_{BaCO3}=100-50,38=49,62\%\)
\(m_{oxit}=m_{hh}-m_{CO2}=7,94-0,06.44=5,3\left(g\right)\)