Đáp án:
\(\begin{array}{l}
{C_\% }{H_2}S{O_4} = 9,3\% \\
{C_\% }FeS{O_4} = 14,42\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{100 \times 20\% }}{{98}} = 0,2\,mol\\
{n_{Fe}} < {n_{{H_2}S{O_4}}} \Rightarrow {H_2}S{O_4} \text{ dư }\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,2 - 0,1 = 0,1\,mol\\
{n_{FeS{O_4}}} = {n_{{H_2}}} = {n_{Fe}} = 0,1\,mol\\
{m_{{\rm{dd}}spu}} = 5,6 + 100 - 0,1 \times 2 = 105,4g\\
{C_\% }{H_2}S{O_4} \text{ dư } = \dfrac{{0,1 \times 98}}{{105,4}} \times 100\% = 9,3\% \\
{C_\% }FeS{O_4} = \dfrac{{0,1 \times 152}}{{105,4}} \times 100\% = 14,42\%
\end{array}\)
