2A+Cl2\(\rightarrow\)2ACl
mCl2=23,4-9,2=14,2(g)
nCl2=\(\frac{14,2}{71}\)=0,2(mol)
\(\rightarrow\)nA=0,2.2=0,4(mol)
MA=\(\frac{9,2}{0,4}\)=23(g/mol)
\(\rightarrow\) A là Natri
\(BTNT.A\Rightarrow n_A=n_{ACl}\Rightarrow\frac{9,2}{A}=\frac{23,4}{A+35,5}\\ \Rightarrow A=23\left(Na\right)\)