\(n_M=\dfrac{1,6}{M_M}\left(mol\right)\)
PTHH: M + Cl2 --to--> MCl2
____\(\dfrac{1,6}{M_M}\)----------->\(\dfrac{1,6}{M_M}\)
=> \(\dfrac{1,6}{M_M}\left(M_M+71\right)=4,44=>M_M=40\left(Ca\right)\)
\(n_{Ca}=\dfrac{1,6}{40}=0,04\left(mol\right)\)
PTHH: Ca + Cl2 --to--> CaCl2
_____0,04->0,04
=> VCl2 = 0,04.22,4 = 0,896(l)