a) \(n_{HCl}=\dfrac{7,3\%.200}{36,5}=0,4\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
Theo PT: `n_{Fe} = n_{H_2} = n_{FeCl_2} = 1/2 n_{HCl} = 0,2 (mol)`
`=> m = 0,2.56 = 11,2 (g)`
b) `m_{dd} = 11,2 + 200 - 0,2.2 = 210,8 (g)`
c) `C\%_{FeCl_2} = (0,2.127)/(210,8) .100\% = 12,05\%`