Question

Answer 1

`n_{H_2} = (3,36)/(22,4) = 0,15 (mol)`

PTHH: `Fe + 2HCl -> FeCl_2 + H_2`

            `Cu + HCl -//->`

Theo PT: `n_{FeCl_2} = n_{H_2} = 0,15 (mol)`

`=> m_{FeCl_2} = 0,15.127 = 19,05 (g)`

`=>` Chọn `D`

Answer 2

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)=n_{Fe}=n_{FeCl_2}\\ m=m_{FeCl_2}=0,15.127=19,05\left(g\right)\\ \Rightarrow Chọn.D\)