\(n\)Fe = \(\dfrac{8,4}{56}\)= 0,15 mol

Fe + 2HCl -----> FeCl\(2\)+H\(2\)

0,15->0,3 ->0,15 -> 0,15 (mol

V\(H2\) = 0,15 . 22,4 = 3,36 l

b, m_{ct HCl} = 0,3 . 36,5 = 10,95 (g)

m_{dd HCl} = \(\dfrac{10,95}{10,95\%}\) = 100 (g)

c, m_{dd sau pu} = 8,4 + 100 - 0,15.2 = 108,1 g

C_{% FeCl2} = \(\dfrac{0,15.127}{108,1}.100\%\)= 1,76%