a, PTHH:Fe+2HCl\(\rightarrow\)FeCl2+H2
b,nFe=\(\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\)(mol)
mHCl=\(\dfrac{C\%.}{100\%}\).mdd=\(\dfrac{14,6.100}{100}=14,6\)(g)
nHCl=\(\dfrac{m}{M}=\dfrac{14,6}{36,5}=0,4\)(mol)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\)
=> Fe hết, HCl dư
Theo PTHH,ta có:
nHCl(PƯ)=2.nFe=2.0,1=0,2(mol)
nHCl(dư)=0,4-0,2=0,2(mol)
mHCl(dư)=n.M=0,2.36,5=7,3(g)
c,Theo PTHH, ta có:
nH2=nFe=0,1(mol)
VH2=n.22,4=0,1.22,4=2,24(l)
d,Theo PTHH, ta có:
nFeCl2=nFe=0,1(mol)
mFeCl2=n.M=0,1.127=12,7(g)