Gọi $n_{Fe} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 56a + 65 = 21,4(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH :
$n_{H_2} = a+ b = \dfrac{7,84}{22,4} = 0,35(2)$
Từ (1)(2) suy ra : a = 0,15 ; b = 0,2
$m_{FeCl_2} = 0,15.127 = 19,05(gam)$
$m_{ZnCl_2} = 0,2.136 = 27,2(gam)$
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a a
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b b
Theo đề, ta có hệ:
56a+65b=21,4 và a+b=0,35
=>a=0,15; b=0,2
=>nFeCl2=0,15mol và nZnCl2=0,2mol
\(m_{FeCl_2}=0.15\cdot\left(56+35.5\cdot2\right)=19.05\left(g\right)\)
\(m_{ZnCl_2}=0.2\cdot\left(65+35.5\cdot2\right)=27.2\left(g\right)\)
Fe+2HCl->FeCl2+H2
Zn+2HCl->ZnCl2+H2
Baot toàn e
m muối=21,4+0,35.2.35,5=46,25g