Fe + 2HCl → FeCl2 + H2 (1)
FeO + 2HCl → FeCl2 + H2O (2)
a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,1\times2=0,2\left(g\right)\)
Theo PT1: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,1\times56=5,6\left(g\right)\)
\(\Rightarrow m_{FeO}=12,8-5,6=7,2\left(g\right)\)
b) Theo PT1: \(n_{HCl}=2n_{Fe}=2\times0,1=0,2\left(mol\right)\)
\(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{FeO}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4\times36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)
c) \(m_{ddFeCl_2}=m_{hh}+m_{ddHCl}-m_{H_2}=12,8+200-0,2=212,6\left(g\right)\)
Theo PT1,2: \(\Sigma n_{FeCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,4=0,2\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,2\times127=25,4\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{212,6}\times100\%=11,95\%\)
