Đáp án:
\(\begin{array}{l}
R:Fe\\
{m_{FeC{l_2}}} = 38,1g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2R + 2nHCl \to 2RC{l_n} + n{H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_R} = \dfrac{{0,3 \times 2}}{n} = \frac{{0,6}}{n}\,mol\\
{M_R} = \dfrac{{16,8}}{{\frac{{0,6}}{n}\,}} = 28n(g/mol)\\
n = 2 \Rightarrow {M_R} = 28 \times 2 = 56(g/mol)\\
\Rightarrow R:\text{ Sắt}(Fe)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,3\,mol\\
{m_{FeC{l_2}}} = 0,3 \times 127 = 38,1g
\end{array}\)