Đáp án:

\(\begin{array}{l}
b)\\
\% {m_{Al}} = 22,44\% \\
\% {m_{Fe}} = 77,56\% \\
c)\\
\% {m_{{H_2}S{O_4}}} = 20\% \\
d)\\
\% {m_{A{l_2}{{(S{O_4})}_3}}} = 40,3\% \\
\% {m_{FeS{O_4}}} = 59,7\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{21,28}}{{22,4}} = 0,95\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
1,5a + b = 0,95\\
27a + 56b = 36,1
\end{array} \right.\\
 \Rightarrow a = 0,3;b = 0,5\\
\% {m_{Al}} = \dfrac{{0,3 \times 27}}{{36,1}} \times 100\%  = 22,44\% \\
\% {m_{Fe}} = 100 - 22,44 = 77,56\% \\
c)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,95\,mol\\
\% {m_{{H_2}S{O_4}}} = \dfrac{{0,95 \times 98}}{{465,5}} \times 100\%  = 20\% \\
d)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,5\,mol\\
\% {m_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,15 \times 342}}{{0,15 \times 342 + 0,5 \times 152}} \times 100\%  = 40,3\% \\
\% {m_{FeS{O_4}}} = 100 - 40,3 = 59,7\% 
\end{array}\)

Đáp án:

Giải thích các bước giải: