Em tham khảo nha:
\(\begin{array}{l}
7)\\
{P_1}:\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
{n_{{H_2}S{O_4}}} = 0,2 \times 1 = 0,2\,mol\\
{n_{FeO}} = {n_{{H_2}S{O_4}}} = 0,2\,mol\\
{P_2}:\\
2FeO + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 4{H_2}O(1)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O(2)\\
{n_{S{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{S{O_2}(1)}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{n_{S{O_2}(2)}} = 0,2 - 0,1 = 0,1\,mol\\
{n_{Cu}} = {n_{S{O_2}(2)}} = 0,1\,mol\\
{m_{hh}} = 0,1 \times 64 + 0,2 \times 72 = 20,8g\\
8)\\
RO + {H_2}S{O_4} \to RS{O_4} + {H_2}O\\
{m_{hh}} = 20 + 180 = 200g\\
{m_{RS{O_4}}} = 200 \times 20\% = 40g\\
{n_{RO}} = {n_{RS{O_4}}} \Leftrightarrow \dfrac{{20}}{{{M_R} + 16}} = \dfrac{{40}}{{{M_R} + 96}}\\
\Rightarrow {M_R} = 64g/mol \Rightarrow R:Cu\\
CTHH:CuO
\end{array}\)
