a, Fe + 2HCl ->FeCl2 + H2
b, nH2= 3.36/22.4=0.15 mol => nFe=0.15mol => nHCl= 0.3
mFe=0.15*56=8.4(g)
c, đổi 100ml= 0.1l => CM HCl= 0.3/0.1 =3M
a.Fe + 2HCl -> FeCl2 + H2
1mol 2mol 1mol 1mol
0.15mol <-0.3mol<-0.15mol <- 0.15mol
b. mFe= 0.15x56=8.4g
c.VHCl=100/1000=0.1l
CMHCl= 0.3/0.1=3M