\(n_{Cl_2}=\dfrac{81,25-28}{71}=0,75\left(mol\right)\)
=> V = 0,75.22,4 = 16,8 (l)
\(n_M=\dfrac{28}{M_M}\left(mol\right)\)
PTHH: 2M + nCl2 --to--> 2MCln
\(\dfrac{28}{M_M}\)-------------->\(\dfrac{28}{M_M}\)
=> \(\dfrac{28}{M_M}\left(M_M+35,5n\right)=81,25\)
=> \(M_M=\dfrac{56}{3}n\left(g/mol\right)\)
- Xét n = 1 => Loại
- Xét n = 2 => Loại
- Xét n = 3 => MM = 56 (g/mol) => Fe