Fe + 2HCl \(\rightarrow\)FeCl2 + H2
nFe=\(\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PTHH ta có:
nFe=nFeCl2=nH2=0,1(mol)
2nFe=nHCl=0,2(mol)
VH2=22,4.0,1=2,24(lít)
\(a)\)\(Fe\left(0,1\right)+2HCl\left(0,2\right)\rightarrow FeCl_2\left(0,1\right)+H_2\left(0,1\right)\)
\(b)\)\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PTHH, ta co: \(n_{HCl}=0,2\left(mol\right)\)
\(n_{FeCl_2}=0,1\left(mol\right)\)
\(n_{H_2}=0,1\left(mol\right)\)
\(c) \)\(V_{H_2}\left(dktc\right)=0,1.22,4=2,24\left(l\right)\)