Đáp án:
a, \(a = {m_{F{\rm{e}}}} = 3,08g\)
b,\({V_{HCl}} = \dfrac{{0,11}}{3} = 0,0367l\)
c,\(C{M_{F{\rm{e}}C{l_2}}} = \dfrac{{0,055}}{{0,0367}} = 1,5M\)
Giải thích các bước giải:
\(\begin{array}{l}
F{\rm{e}} + 2HCl \to F{\rm{e}}C{l_2} + {H_2}\\
{n_{{H_2}}} = 0,055mol\\
\to {n_{F{\rm{e}}}} = {n_{{H_2}}} = 0,055mol\\
\to a = {m_{F{\rm{e}}}} = 3,08g\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,11mol\\
\to {V_{HCl}} = \dfrac{{0,11}}{3} = 0,0367l\\
{n_{F{\rm{eC}}{{\rm{l}}_2}}} = {n_{{H_2}}} = 0,055mol\\
\to C{M_{F{\rm{e}}C{l_2}}} = \dfrac{{0,055}}{{0,0367}} = 1,5M
\end{array}\)
