FeO + H2 \(\underrightarrow{to}\) Fe + H2O (1)
\(n_{FeO}=\frac{14,4}{72}=0,2\left(mol\right)\)
\(n_{H_2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PT1: \(n_{FeO}=n_{H_2}\)
Theo bài: \(n_{FeO}=4n_{H_2}\)
Vì \(4>1\) ⇒ FeO dư
Chất rắn A: FeO dư và Fe
Fe + 2HCl → FeCl2 + H2 (2)
FeO + 2HCl → FeCl2 + H2O (3)
a) Theo PT1: \(n_{Fe}=n_{H_2}=0,05\left(mol\right)\)
Theo PT1: \(n_{FeO}pư=n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow n_{FeO}dư=n_{FeO}\left(3\right)=0,2-0,05=0,15\left(mol\right)\)
Theo pT2: \(n_{HCl}=2n_{Fe}=2\times0,05=0,1\left(mol\right)\)
Theo PT3: \(n_{HCl}=2n_{FeO}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,1+0,3=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4\times36,5=14,6\left(g\right)\)
\(\Rightarrow m=m_{ddHCl}=\frac{14,6}{10\%}=146\left(g\right)\)
b) Theo PT2: \(n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\)
Theo Pt3: \(n_{FeCl_2}=n_{FeO}=0,15\left(mol\right)\)
\(\Rightarrow\Sigma n_{FeCl_2}=0,05+0,15=0,2\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,2\times127=25,4\left(g\right)\)
\(m_{Fe}=0,05\times56=2,8\left(g\right)\)
\(m_{FeO}dư=0,15\times72=10,8\left(g\right)\)
Theo PT2: \(n_{H_2}=n_{Fe}=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,05\times2=0,1\left(g\right)\)
Ta có: \(m_{ddX}=2,8+10,8+146-0,1=159,5\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\frac{25,4}{159,5}\times100\%=15,92\%\)
