PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
Ta có: \(n_{CaCO_3}=\dfrac{2,1.10^{24}}{6.10^{23}}=3,5\left(mol\right)\)
a, Theo PT: \(n_{CaO}=n_{CO_2}=n_{CaCO_3}=3,5\left(mol\right)\)
\(\Rightarrow A_{CaO}=A_{CO_2}=3,5.6.10^{23}=2,1.10^{24}\) (phân tử)
b, \(m_{CaO}=3,5.56=196\left(g\right)\)
\(m_{CO_2}=3,5.44=154\left(g\right)\)
c, \(V_{CO_2}=3,5.24,79=86,765\left(l\right)\)