Đáp án:
\(C{\% _{{H_2}S{O_4}}}dư= \dfrac{{1,9 \times 98}}{{205,4}} \times 100\% = 90,65\% \)
\(C{\% _{FeS{O_4}}} = \dfrac{{0,1 \times 152}}{{205,4}} \times 100\% = 7,4\% \)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 98}}{{100}} = 196g\\
\to {n_{{H_2}S{O_4}}} = 2mol\\
\to {n_{Fe}} < {n_{{H_2}S{O_4}}}
\end{array}\)
Suy ra sau phản ứng có \({H_2}S{O_4}\) dư
\(\begin{array}{l}
\to {n_{{H_2}S{O_4}}}dư= 2 - 0,1 = 1,9mol\\
\to {n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
\to {n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
{m_{{\rm{dd}}}} = {m_{Fe}} + {m_{{H_2}S{O_4}{\rm{dd}}}} - {m_{{H_2}}} = 205,4g\\
\to C{\% _{{H_2}S{O_4}}}dư= \dfrac{{1,9 \times 98}}{{205,4}} \times 100\% = 90,65\% \\
\to C{\% _{FeS{O_4}}} = \dfrac{{0,1 \times 152}}{{205,4}} \times 100\% = 7,4\%
\end{array}\)
