n\(_{Zn}\)= \(\dfrac{3,25}{65}\)= 0,05 (mol)
PTHH: Zn + 2HCl ----> ZnCl2 + H2\(\uparrow\)
mol: _0,05->0,1--------> 0,05--->0,05
V\(_{H_2}\)= 0,05 . 22,4 = 1,12 (lít)
m\(_{HCl}\) = 0,1 . 36,5 = 3,65 (g)
m\(_{ddHCl}\) = \(\dfrac{3,65.100}{14,6}\)= 25 (g)
m\(_{ZnCl_2}\)= 0,05 . 136 = 6,8 (g)
m\(_{ddZnCl_2}\)= 3,25 + 25 = 28,25 (g)
C%ZnCl2 = \(\dfrac{6,8}{28,25}\).100% = 24,07%