Đáp án:
\(\begin{array}{l}
a)\\
A:Al\\
b)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 45g\\
c)\\
{C_\% }AlC{l_3} = 76,72\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2A + 3{H_2}S{O_4} \to {A_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45\,mol\\
{n_A} = 0,45 \times \frac{2}{3} = 0,3\,mol\\
{M_A} = \dfrac{{8,1}}{{0,3}} = 27\,g/mol \Rightarrow A:Al\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,45\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,45 \times 98}}{{98\% }} = 45g\\
c)\\
{m_{{\rm{dd}}B}} = 8,1 + 45 - 0,45 \times 2 = 52,2g\\
{C_\% }AlC{l_3} = \dfrac{{0,3 \times 133,5}}{{52,2}} \times 100\% = 76,72\%
\end{array}\)