\(n_{Na_2CO_3\cdot10H_2O}=\dfrac{143}{106+10\cdot18}=0.5\left(mol\right)\)
\(\Rightarrow n_{Na_2CO_3}=n_{Na_2CO_3\cdot10H_2O}=0.5\left(mol\right)\)
\(m_{Na_2CO_3}=0.5\cdot106=53\left(g\right)\)
\(S_{Na_2CO_3}=\dfrac{53}{160}\cdot100=33.125\left(g\right)\)