Đáp án:
\(\begin{array}{l}
a)\\
M:(Fe)\\
b)\\
{m_{FeC{l_2}}} = 12,7g\\
c)\\
C{\% _{FeC{l_2}}} = 16,2\% \\
C{\% _{HCl}} = 13,97\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
M + 2HCl \to MC{l_2} + {H_2}\\
{n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1mol\\
\Rightarrow {n_M} = {n_{{H_2}}} = 0,1mol\\
{M_M} = \dfrac{{5,6}}{{0,1}} = 56dvC\\
\Rightarrow M:\text{Sắt}(Fe)\\
b)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
c)\\
{m_{HCl}} = \dfrac{{73 \times 25}}{{100}} = 18,25g\\
{n_{HCl}} = \dfrac{{18,25}}{{36,5}} = 0,5mol\\
{n_{HC{l_d}}} = {n_{HCl}} - 2{n_{{H_2}}} = 0,5 - 0,2 = 0,3mol\\
{m_{HC{l_d}}} = 0,3 \times 36,5 = 10,95g\\
{m_{{H_2}}} = 0,1 \times 2 = 0,2g\\
{m_{{\rm{ddspu}}}} = 5,6 + 73 - 0,2 = 78,4g\\
C{\% _{FeC{l_2}}} = \dfrac{{12,7}}{{78,4}} \times 100\% = 16,2\% \\
C{\% _{HCl}} = \dfrac{{10,95}}{{78,4}} \times 100\% = 13,97\%
\end{array}\)
