a) Fe+2HCl--->FeCl2+H2
b) n Fe=5,6/56=0,1(mol)
Theo pthh
n H2=n Fe=0,1(mol)
V\(_{H2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
c) Theo pthh
n\(_{FeCl2}=n_{Fe}=0,1\left(mol\right)\)
m\(_{FeCl2}=0,1.127=12,7\left(g\right)\)
a, \(Fe+HCl--->FeCl2+H2\)
b,
Ta có :
nFe = 0,1 (mol)
=> nH2 = nFe = 0,1 (mol)
=> V H2 = 0,1 .22,4 = 2,24l
c, =>nFeCl2 = nFe = 0,1 (mol)
=>mFeCl2 = 12,7(g)