Question

Answer 1

`-` Cốc `A:`

`n_{Fe} = (16,8)/(56) = 0,3(mol)`

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

Theo PT: `n_{H_2} = n_{Fe} = 0,3(mol)`

`-> m_{tăng} = 16,8 - 0,3.2 = 16,2(g)(1)`

`-` Cốc `B:`

`n_{Al} = m/(27) (mol)`

`PTHH: 2Al + 6HCl -> 2AlCl_3 + 3H_2`

Theo PT: `n_{H_2} = 3/2 n_{Al} = 3/2 . m/(27) = m/(18) (mol)`

`-> m_{tăng} = m - 2 . m/(18) = (8m)/9 (g)`

`-> (8m)/9 = 16,2`

`<=> m = 18,225(g)`