$n_{H_{2}}$ = 5,6/22,4 = 0,25 mol
Gọi x = $n_{Fe}$; y = $n_{Zn}$
a)
PT: Fe + 2HCl ---> Fe$Cl_{2}$ + $H_{2}$
PƯ: x 2x x (mol)
PT: Zn + 2HCl ---> Zn$Cl_{2}$ + $H_{2}$
PƯ: y 2y y (mol)
b) Ta có:
$\left \{ {{56x + 65y = 14,9} \atop {x+y= 0,25}} \right.$
=> $\left \{ {{x=0,15} \atop {y=0,1}} \right.$
$m_{Fe}$ = 0,15 . 56 = 8,4g
$m_{Zn}$ = 0,1 . 65 = 6,5g
c) $m_{HCl}$ = (0,15 . 2 + 0,1 . 2). 36,5 = 18,25 g