Đáp án:
\(\begin{array}{l}
R:Al\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 51,3g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2R + n{H_2}S{O_4} \to {R_2}{(S{O_4})_n} + n{H_2}\\
{n_{{H_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45\,mol\\
{n_R} = \dfrac{{0,45 \times 2}}{n} = \dfrac{{0,9}}{n}\,mol\\
{M_R} = \dfrac{{8,1}}{{\dfrac{{0,9}}{n}\,}} = 9n(g/mol)\\
n = 3 \Rightarrow {M_R} = 9 \times 3 = 27(g/mol)\\
\Rightarrow R:\text{ Nhôm}(Al)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,45}}{3} = 0,15\,mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,15 \times 342 = 51,3g
\end{array}\)
