Em tham khảo nha :
\(\begin{array}{l}
7)\\
Fe + 2HCl \to FeC{l_2} + {H_2}(1)\\
2Y + 2xHCl \to 2YC{l_x} + x{H_2}(2)\\
{n_{{H_2}}} = \dfrac{{7,392}}{{22,4}} = 0,33mol\\
{n_{Fe}} = \dfrac{{8,4}}{{56}} = 0,15mol\\
{n_{{H_2}(1)}} = {n_{Fe}} = 0,15mol\\
{n_{{H_2}(2)}} = 0,33 - 0,15 = 0,18mol\\
{n_Y} = \dfrac{2}{x}{n_{{H_2}}} = \dfrac{{0,36}}{x}mol\\
{M_Y} = 3,24:\dfrac{{0,36}}{x} = 9x\,dvC\\
x = 3 \Rightarrow {M_Y} = 27dvC\\
Y:\text{Nhôm}(Al)\\
8)\\
Y + 2HCl \to YC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{m_Y} = 12 - 6,4 = 5,6g\\
{M_Y} = \dfrac{{5,6}}{{0,1}} = 56dvC\\
Y:\text{Sắt}(Fe)\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g
\end{array}\)