A + Cl2 \(\underrightarrow{t^o}\) ACl2
1 ....... 1 .......... 1 .... (mol)
0,1.. 0,1......0,1... (mol)
a) \(^nCl_2=\dfrac{V_{đktc}}{22,4}=\dfrac{2,24}{22,4}-0,1\left(mol\right)\)
MA = \(\dfrac{m}{n}=\dfrac{2,4}{0,1}=24\) (g/mol)
Vậy kim loại A là Mg.
b) mMgCl2 = n.M = 0,1.95 = 9,5 (g)