Em tham khảo nha :
\(\begin{array}{l}
1)\\
xR + y{H_2}S{O_4} \to {R_x}{(S{O_4})_y} + y{H_2}\\
{n_{{H_2}}} = \dfrac{{50,4}}{{22,4}} = 2,25mol\\
{n_R} = \dfrac{x}{y}{n_{{H_2}}} = \dfrac{{2,25x}}{y}mol\\
{M_R} = 146,25:\dfrac{{2,25x}}{y} = 65\frac{y}{x}\\
x = 2;y = 2 \Rightarrow {M_R} = 65dvC\\
\Rightarrow R:\text{Kẽm}(Zn)\\
b)\\
{n_{ZnS{O_4}}} = {n_{{H_2}}} = 2,25mol\\
{m_{ZnS{O_4}}} = 2,25 \times 161 = 362,25g\\
{m_{{\rm{dd}}spu}} = 146,25 + 758,25 - 2,25 \times 2 = 900g\\
C{\% _{ZnS{O_4}}} = \dfrac{{362,25}}{{900}} \times 100\% = 40,25\%
\end{array}\)