a, PT: \(Na_2CO_3+Ca\left(OH\right)_2\rightarrow2NaOH+CaCO_{3\downarrow}\)
Ta có: \(n_{NaOH}=0,05.2=0,1\left(mol\right)\)
Theo PT: \(n_{CaCO_3}=\dfrac{1}{2}n_{NaOH}=0,05\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,05.100=5\left(g\right)\)
b, Theo PT: \(n_{Na_2CO_3}=\dfrac{1}{2}n_{NaOH}=0,05\left(mol\right)\)
\(\Rightarrow C\%_{Na_2CO_3}=\dfrac{0,05.106}{53}.100\%=10\%\)