Question

Answer 1

1: Fe+CuCl2->FeCl2+Cu

nCu=3,2/64=0,05(mol)

=>n Fe=0,05(mol)

m=0,05*56=2,8(g)

nFeCl2=nFe=0,05(mol)

mFeCl2=0,05(56+2*35,5)=6,35(g)

 

Answer 2

2

Gọi `n_(Fe)=x(mol)`

\(Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\)

x---------2x------------------------->2x

Có: \(m_{kim.loại.tăng}=108.2x-56x=5,2\left(g\right)\)

\(\Rightarrow x=0,0325\left(mol\right)\)

\(\Rightarrow m_{AgNO_3}=2.0,0325.170=11,05\left(g\right)\)