Đáp án:
\(\begin{array}{l}
b,\\
C{\% _{{H_2}S{O_4}}} = 9,8\% \\
C{\% _{FeS{O_4}}} = 14,4\% \\
c,\\
{m_{Fe{{(OH)}_2}}} = 18g\\
C{\% _{N{a_2}S{O_4}}} = 8,05\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
a,
\end{array}\)
Hiện tượng: Sắt (Fe) tan dần trong dung dịch và sinh ra khí Hidro (H2) làm sủi bọt khí.
\(\begin{array}{l}
b,\\
{n_{Fe}} = 0,2mol\\
\to {n_{{H_2}S{O_4}}} = {n_{FeS{O_4}}} = {n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
\to {m_{{H_2}S{O_4}}} = 19,6g\\
\to {m_{FeS{O_4}}} = 30,4g\\
\to {m_{{H_2}}} = 0,4g
\end{array}\)
\(\begin{array}{l}
{m_{{\rm{dd}}}} = {m_{Fe}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 210,8g\\
\to C{\% _{{H_2}S{O_4}}} = \dfrac{{19,6}}{{200}} \times 100\% = 9,8\% \\
\to C{\% _{FeS{O_4}}} = \dfrac{{30,4}}{{210,8}} \times 100\% = 14,4\%
\end{array}\)
\(\begin{array}{l}
c,\\
FeS{O_4} + 2NaOH \to Fe{(OH)_2} + N{a_2}S{O_4}\\
{n_{NaOH}} = 2{n_{FeS{O_4}}} = 0,4mol\\
\to {m_{NaOH}} = 16g\\
\to {m_{{\rm{dd}}NaOH}} = \dfrac{{16}}{{10\% }} \times 100\% = 160g
\end{array}\)
\(\begin{array}{l}
{n_{Fe{{(OH)}_2}}} = {n_{N{a_2}S{O_4}}} = {n_{FeS{O_4}}} = 0,2mol\\
\to {m_{Fe{{(OH)}_2}}} = 18g\\
\to {m_{N{a_2}S{O_4}}} = 28,4g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}FeS{O_4}}} + {m_{{\rm{dd}}NaOH}} - {m_{Fe{{(OH)}_2}}} = 352,8g\\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{28,4}}{{352,8}} \times 100\% = 8,05\%
\end{array}\)
