Đáp án:
\(\begin{array}{l}
a)\\
A:\text{Nhôm}(Al)\\
b)\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = 16,7\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2A + 3{H_2}S{O_4} \to {A_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6mol\\
{n_A} = \dfrac{2}{3}{n_{{H_2}}} = 0,4mol\\
{M_A} = \dfrac{{10,8}}{{0,4}} = 27dvC\\
A:\text{Nhôm}(Al)\\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{{H_2}}}}}{3} = 0,2mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,2 \times 342 = 68,4g\\
{m_{ddspu}} = 10,8 + 400 - 0,6 \times 2 = 409,6g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{68,4}}{{409,6}} \times 100\% = 16,7\%
\end{array}\)