\(n_{CaO\left(bđ\right)}=\dfrac{325.1000}{56}=\dfrac{40625}{7}\left(mol\right)\)
PTHH: \(CaO+CO_2\rightarrow CaCO_3\)
Theo PTHH: \(n_{CaCO_3\left(tt\right)}=n_{CaO\left(bđ\right)}=\dfrac{40625}{7}\left(mol\right)\)
=> \(H\%=\dfrac{475.1000}{\dfrac{40625}{7}.100}.100\%=81,85\%\)