\(Fe + 2HCl \to FeCl_2 + H_2\)
Theo PTHH :
\(n_{HCl} = 2n_{Fe} = 2.\dfrac{11,2}{56} = 0,4(mol)\)
\(NaOH + HCl \to NaCl + H_2O\)
Ta có :
\(n_{NaOH} = n_{HCl} = 0,4(mol)\\ \Rightarrow m_{dd\ NaOH} = \dfrac{0,4.40}{20\%} = 80(gam)\\ \Rightarrow V_{dd\ NaOH} = \dfrac{80}{1,12} = 71,43(ml)\)