Đáp án:
\(\begin{array}{l}
a)\\
{m_{Fe{{(OH)}_2}}} = 9g\\
b)\\
C{\% _{{K_2}S{O_4}}} = 5,979\% \\
c)\\
{m_{FeO}} = 7,2g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
FeS{O_4} + 2KOH \to Fe{(OH)_2} + {K_2}S{O_4}\\
a)\\
{m_{FeS{O_4}}} = \dfrac{{100 \times 15,2\% }}{{100\% }} = 15,2g\\
{n_{FeS{O_4}}} = \dfrac{m}{M} = \dfrac{{15,2}}{{152}} = 0,1mol\\
\Rightarrow {n_{Fe{{(OH)}_2}}} = {n_{FeS{O_4}}} = 0,1mol\\
{m_{Fe{{(OH)}_2}}} = n \times M = 0,1 \times 90 = 9g\\
b)\\
{n_{{K_2}S{O_4}}} = {n_{FeS{O_4}}} = 0,1mol\\
{m_{{K_2}S{O_4}}} = n \times M = 0,1 \times 174 = 17,4g\\
{m_{{\rm{dd}}spu}} = {m_{{\rm{dd}}FeS{O_4}}} + {m_{{\rm{dd}}KOH}} - {m_{Fe{{(OH)}_2}}} = 100 + 200 - 9 = 291g\\
C{\% _{{K_2}S{O_4}}} = \dfrac{{{m_{{K_2}S{O_4}}}}}{{{m_{{\rm{dd}}spu}}}} \times 100\% = \dfrac{{17,4}}{{291}} \times 100\% = 5,979\% \\
c)\\
Fe{(OH)_2} \to FeO + {H_2}O\\
{n_{FeO}} = {n_{Fe{{(OH)}_2}}} = 0,1mol\\
{m_{FeO}} = n \times M = 0,1 \times 72 = 7,2g
\end{array}\)
