PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
Ta có: \(n_{CaO}=\dfrac{7}{56}=0,125\left(mol\right)\)
a, Theo PT: \(n_{CaCO_3}=n_{CaO}=0,125\left(mol\right)\)
\(\Rightarrow A_{CaCO_3}=0,125.6,10^{23}=0,75.10^{23}\) (phân tử)
\(m_{CaCO_3}=0,125.100=12,5\left(g\right)\)
b, \(n_{CO_2}=n_{CaO}=0,125\left(mol\right)\)
\(A_{CO_2}=0,125.6.10^{23}=0,75.10^{23}\) (phân tử)
\(m_{CO_2}=0,125.44=5,5\left(g\right)\)
c, \(V_{CO_2}=0,125.24,79=3,09875\left(l\right)\)