Đáp án:
\(\begin{array}{l}
a)\\
\% Fe = 56\% \\
\% Ag = 44\% \\
b)\\
C{\% _{{H_2}S{O_4}}} = 6,53\% \\
c)\\
{m_{FeS{O_4}}} = 30,4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
\% Fe = \dfrac{{11,2}}{{20}} \times 100\% = 56\% \\
\% Ag = 100 - 56 = 44\% \\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = 0,2 \times 98 = 19,6g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{19,6}}{{300}} \times 100\% = 6,53\% \\
c)\\
{n_{FeS{O_4}}} = {n_{{H_2}}} = 0,2mol\\
{m_{FeS{O_4}}} = 0,2 \times 152 = 30,4g
\end{array}\)
