Ta có: \(n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Bảo toàn Cacbon: \(n_{CaCO_3}=n_{CO_2}=0,02\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,02\cdot100}{5}\cdot100\%=40\%\\\%m_{CaSO_4}=60\%\end{matrix}\right.\)