· Phần 1:

$NaHCO_3+HCl\xrightarrow\ NaCl+CO_2+H_2O\\ n_{CO_2}=n_{NaHCO_3}=\dfrac{4,2}{84}=0,05(mol)\\ m_{1\ tang}=4,2-0,05.44=2(g)$

· Phần 2:

$Fe+2HCl\xrightarrow\ FeCl_2+H_2\\ n_{Fe(pu)}=n_{H_2}=a(mol)\\ m_{2\ tang}=56a-2a(g)$

Để cân thăng bằng:

$56a-2a=2\xrightarrow\ a=\dfrac{1}{27}(mol)\\ m_{Fe}=\dfrac{1}{27}.56=2,074(g)$

Cốc 1:

`n_{NaHCO_3}={4,2}/{84}=0,05(mol)`

`NaHCO_3+HCl->NaCl+CO_2+H_2O`

`0,05->0,05->0,05->0,05(mol)`

`->\Delta m_1=4,2-0,05.44=2(g)`

Cốc 2:

`Fe+2HCl->FeCl_2+H_2`

Theo pt:

`n_{Fe}=n_{H_2}=x(mol)`

`->\Delta m_2=56x-2x=\Delta m_1=2`

`->x=1/{27}(mol)`

`->m_{Fe}=1/{27}.56\approx 2,074(g)`