Question

Answer 1

nFe = 5,6/56 = 0,1 (mol)

Fe + 2HCl --> FeCl2 + H2

0,1     0,2        0,1       0,1

=> VH2 = 0,1.22,4 = 2,24 (l)

b) mFeCl2 = 0,1.127= 12,7g

c) VHCl = 100ml = 0,1 (l)

CM HCl = 0,2/0,1 = 2M

Answer 2

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,1---->0,2------->0,1---->0,1

=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1.127=12,7\left(g\right)\\C_{M\left(HCl\right)}=\dfrac{0,2}{0,1}=2M\end{matrix}\right.\)