`PTHH:Fe+2HCl→FeCl_2+H_2↑`
a)Ta có: `n_(Fe)=m/M=(2,8)/56=0,05(mol)`
Theo `PTHH`,ta thấy: `n_(Fe)=n_(H_2)=0,05(mol)`
`⇒V_(H_2)=n.22,4=0,05.22,4=1,12(l)`
b)Theo `PTHH`,ta thấy: `n_(HCl)=2n_(Fe)=2.0,05=0,1(mol)`
`⇒m_(HCl)=n.M=0,1.(1+35,5)=3,65(g)`
`⇒m_(ddHCl)=100.(3,65)/25=14,6(g)`
c)Theo `PTHH`,ta thấy: `n_(Fe)=n_(FeCl_2)=0,05(mol)`
`⇒m_(ctFeCl_2)=n.M=0,05.127=6,35(g)`
Theo Nguyên tắc bảo toàn khối lượng,ta có:
`m_(Fe)+m_(ddHCl)=m_(ddFeCl_2)+m_(H_2)`
`⇒m_(ddFeCl_2)=(0,05 . 56) + 14,6 - (0,05 . 2) = 17,3(g)`
`⇒C%ddFeCl_2=100.(6,35)/(17,3)≈36,71%`
Xin hay nhất =_=