a) \(n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
PTHH: CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O
0,02<-----0,04<------------------0,02
=> \(C_{M\left(HCl\right)}=\dfrac{0,04}{0,02}=2M\)
b) \(\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,02.100}{4}.100\%=50\%\\\%m_{NaCl}=100\%-50\%=50\%\end{matrix}\right.\)