Bài 3
`a)`
PTHH
`Fe + 2HCl→FeCl_2 + H_2↑`
`b)`
`n_(H_2)=(3,36)/(22,4)=0.15(mol)`
Theo PTHH
`n_(H_2)=n_(Fe)=0.15(mol)`
`⇒m_(Fe)=0.15.56=8,4(g)`
`⇒%m_(Fe)=(8,4)/(12).100=70%`
`⇒%m_(Cu)=100%-70%=30%`
`c)`
Theo PTHH
`n_(HCl)=n_(H_2)=0.15.2=0,3(mol)`
`⇒m_(HCl)=0,3.36,5=10,95(g)`
`⇒m_(dd_(HCl))=(10,95)/(10%)=109,5(g)`
`d)`
Theo PTHH
`n_(FeCl_2)=n_(Fe)=0.15(mol)`
`m_(FeCl_2)=0,15.127=19,5(g)`
Theo PTHH
`n_(H_2)=n_(HCl)=0.15(mol)`
`m_(H_2)=0,15.2=0,3(g)`
`⇒m_(dd_(FeCl_2))=109,5+8,4-0,3=117.6(g)`
`⇒C%_(FeCl_2)=(19,5)/(117,6).100%=16,58%`