Fe + 2HCl → FeCl2 + H2 (1)
\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
a) Theo PT1: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)
b) Theo pT1: \(n_{HCl}=2n_{Fe}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4\times36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{14,6}{7,3\%}=200\left(g\right)\)
c) \(m_{H_2}=0,2\times2=0,4\left(g\right)\)
Ta có: \(m_{dd}saupư=11,2+200-0,4=210,8\left(g\right)\)
Theo pT1: \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,2\times127=25,4\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\frac{25,4}{210,8}\times100\%=12,05\%\)
