PTPU
2A+ Cl2\(\xrightarrow[]{to}\) 2ACl
ADĐLBTKL có
mCl2= mmuối- mkl
= 23,4- 9,2= 14,2( g)
\(\Rightarrow\) nCl2= \(\dfrac{14,2}{71}\)= 0,2( mol)
theo PTPU có: nA= 2nCl2= 0,4( mol)
\(\Rightarrow\) MA= \(\dfrac{m}{n}\)= \(\dfrac{9,2}{0,4}\)= 23
vậy A là natri( Na)