\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
\(0.1...............0.1..............0.1\)
\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{dd_{Ca\left(OH\right)_2}}=\dfrac{0.1\cdot74}{8.55\%}=86.55\left(g\right)\)