Question

Answer 1

\(1.Fe+2HCl\rightarrow FeCl_2+H_2\\ 2.n_{HCl}=0,4.1,5=0,6mol\\ n_{H_2}=n_{Fe}=n_{FeCl_2}=0,6:2=0,3mol\\ m_{Fe}=0,3.56=16,8g\\ 3.V_{H_2}=0,3.22,4=6,72l\\ 4.C_{M_{FeCl_2}}=\dfrac{0,3}{0,4}=0,75M\)

Answer 2

\(n_{HCl}=1,5.0,4=0,6\left(mol\right)\)

1) PTHH : \(Fe+2HCl\rightarrow FeCl_2+H_2\)

2) Theo Pt : \(n_{H2}=n_{Fe}=n_{FeCl2}=n_{H2}=0,6\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,6.56=33,6\left(g\right)\)

3) \(V_{H2\left(dktc\right)}=0,6.22,4=13,44\left(l\right)\)

3) \(C_{MFeCl2}=\dfrac{0,6}{0,4}=1,5\left(M\right)\)