Đáp án:
\(\begin{array}{l}
b)\\
m = 22g\\
c)\\
\% {m_{Fe}} = 50,91\% \\
\% {m_{Al}} = 49,09\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
{n_A} = 33,33\% \times 21 = 7\\
2{p_A} + {n_A} = 21 \Rightarrow {p_A} = {e_A} = 7\\
\Rightarrow A:Nito(N)\\
2)\\
a)\\
{P_1}:\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}(1)\\
{P_2}:\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}(2)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}(3)\\
b)\\
{n_{NaOH}} = 0,2 \times 1 = 0,2\,mol\\
{n_{Al}} = {n_{NaOH}} = 0,2\,mol\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{{H_2}(2)}} = 0,2 \times \dfrac{3}{2} = 0,3\,mol\\
{n_{{H_2}(3)}} = 0,4 - 0,3 = 0,1\,mol\\
{n_{Fe}} = {n_{{H_2}(3)}} = 0,1\,mol\\
m = 0,2 \times 2 \times 27 + 0,1 \times 2 \times 56 = 22g\\
c)\\
\% {m_{Fe}} = \dfrac{{0,2 \times 56}}{{22}} \times 100\% = 50,91\% \\
\% {m_{Al}} = 100 - 50,91 = 49,09\%
\end{array}\)
