\(m_{Na}=\dfrac{43,4.106}{100}=46\left(g\right)=>n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)
\(m_C=\dfrac{11,3.106}{100}=12\left(g\right)=>n_C=\dfrac{12}{12}=1\left(mol\right)\)
\(m_O=\dfrac{45,3.106}{100}=48\left(g\right)=>n_O=\dfrac{48}{16}=3\left(mol\right)\)
=> CTHH: Na2CO3