ms = 227 g , c% = 28%
stim ca c% = mdx100/ms
=> md = msxc%/100
= 227x28/100 = 63,56 g FeCl2
md g 63,56 g
Fe + 2HCl --> FeCl2 + H2
2x36,5 127
=> md.HCl = 36,53 g HCl
ms.HCl = md + m.apa din sol finala
m.apa din sol finala = 227 g - 63,56 = 163,44 g
=> ms.HCl = 36,53 + 163,44 = 199,97 g
=> c%.HCl = 36,53x100/199,97 = 18,28%
=> corecta a)