a)
`PTHH:2Fe+3Cl_2 \overset{t^0}\to 2FeCl_3`
Ta có: `n_(Fe)=m/M=(5,6)/56=0,1(mol)`
b)
Theo `PTHH`: `n_(FeCl_3)=n_(Fe)=0,1(mol)`
`=>m_(FeCl_3)=n.M=0,1.162,5=16,25(g)`
c)
Theo `PTHH`: `n_(Cl_2)=3/2.n_(Fe)=3/2.0,1=0,15(mol)`
`=>V_(Cl_2)=n.22,4=0,15.22,4=3,36(l)`