\(M_A=\dfrac{21,2}{0,2}=106\left(\dfrac{g}{mol}\right)\)
\(m_{Na}=\dfrac{106.43,4}{100}=46g\)
\(m_C=\dfrac{106.11,32}{100}=12g\)
\(m_O=106-46-12=48g\)
\(n_{Na}=\dfrac{46}{23}=2mol\)
\(n_C=\dfrac{12}{12}=1mol\)
\(n_O=\dfrac{48}{16}=3mol\)
\(\Rightarrow CTHH:Na_2CO_3\)
\(\Rightarrow\) Đáp án D