Fe+2HCl->FeCl2+H2
0,015-------------------0,015 mol
n H2=0,336\22,4=0,015 mol
=>m Fe=0,015.56=0,84g
PTHH
Fe + 2HCL --> Fecl2 +H2
TBR, ta có : nH2 = 0,336 / 22,4 = .............(mol) (tự tính)
Theo PTHH, nFe = nH2 = .........(tính trên kia)
=> m Fe = nFe . 56 = (g)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,015.56=0,84\left(g\right)\)