Em tham khảo nha :
\(\begin{array}{l}
a)\\
2X + 2aHCl \to 2XC{l_a} + a{H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_X} = \dfrac{2}{a}{n_{{H_2}}} = \dfrac{{0,6}}{a}mol\\
{M_X} = 16,8:\dfrac{{0,6}}{a} = 28a\\
a = 2 \Rightarrow {M_X} = 56dvC\\
\Rightarrow X:\text{Sắt}(Fe)\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6mol\\
{m_{HCl}} = 0,6 \times 36,5 = 21,9g\\
C{\% _{HCl}} = \dfrac{{21,9}}{{200}} \times 100\% = 10,95\% \\
c)\\
{n_{FeS{O_4}}} = {n_{{H_2}}} = 0,3mol\\
{m_{FeC{l_2}}} = 0,3 \times 127 = 38,1g\\
{m_{ddspu}} = 16,8 + 200 - 0,3 \times 2 = 216,2g\\
C{\% _{FeS{O_4}}} = \dfrac{{38,1}}{{216,2}} \times 100\% = 17,6\%
\end{array}\)
