Đáp án:

\(\begin{array}{l}
A:Fe\\
FeS{O_4}.7{H_2}O\\
{C_\% }FeS{O_4} = 14,64\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
A + {H_2}S{O_4} \to AS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,4}}{{22,4}} = 0,375\,mol\\
{n_A} = {n_{{H_2}}} = 0,375\,mol\\
{M_A} = \dfrac{{21}}{{0,375}} = 56g/mol \Rightarrow A:Fe\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,375\,mol\\
 \Rightarrow 0,375 \times (152 + 18n) = 104,25\\
 \Rightarrow n = 7 \Rightarrow CT:FeS{O_4}.7{H_2}O\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,35\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,35 \times 98}}{{10\% }} = 343g\\
{m_{{\rm{dd}}B}} = 21 + 343 - 0,35 \times 2 = 363,3g\\
{C_\% }FeS{O_4} = \dfrac{{0,35 \times 152}}{{363,3}} \times 100\%  = 14,64\% 
\end{array}\)