a) \(PTHH:CaCO_3\underrightarrow{t^o}CaO+CO_2\)
b) \(m_{CaCO_3}=120.90\%=108\left(g\right)\)
\(n_{CaCO_3}=\frac{m_{CaCO_3}}{M_{CaCO_3}}=\frac{108}{100}=1,08\left(mol\right)\)
\(Theo\) \(PTHH,\) \(ta có:\)
\(n_{CaO}=n_{CaCO_3}=1,08\left(mol\right)\)
\(m_{CaO}=n_{CaO}.M_{CaO}=1,08.56=60,48\left(g\right)\)