\(PTHH:CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
\(n_{Ca\left(OH\right)_2}=\dfrac{m}{M}=\dfrac{37}{74}=0,5\left(mol\right)\)
\(Theo.PTHH\Rightarrow n_{CO_2}=n_{CaCO_3}=n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)
\(m_{CO_2}=n.M=0,5.44=22\left(g\right)\)