a, Ta có: \(n_{CaO}=\dfrac{9,72}{56}=\dfrac{243}{1400}\left(mol\right)\)
PT: \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaO}=\dfrac{243}{1400}\left(mol\right)\)
\(\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{\dfrac{243}{1400}}{0,7}\approx0,25\left(M\right)\)
b, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Theo PT: \(n_{CO_2}=n_{Ca\left(OH\right)_2}=\dfrac{243}{1400}\left(mol\right)\)
\(\Rightarrow V_{CO_2}=\dfrac{243}{1400}.22,4=3,888\left(l\right)\)