Đáp án đúng:
Giải chi tiết:\(X\left\{ \matrix{ Cu \hfill \cr Fe \hfill \cr} \right.\buildrel { + {H_2}S{O_4}:\,0,5mol} \over \longrightarrow \left\{ \matrix{ ran\,Z:\,Cu(0,09875) \hfill \cr {\rm{dd}}\,Y\buildrel { + {H_2}O} \over \longrightarrow \underbrace {{\rm{dd}}\,{Y_1}}_{600ml} \to \left[ \matrix{ \buildrel {co\,can} \over \longrightarrow ran\,T \hfill \cr \buildrel {{\rm{dd}}\,U} \over \longrightarrow \hfill \cr} \right. \hfill \cr} \right.\)
a.Gọi a, b lần lượt là số mol của Cu và Fe3O4
\( \to \left\{ \matrix{ 64a + 232b = 30 \hfill \cr a - b = 0,09875 \hfill \cr} \right. \to \left\{ \matrix{ a = 0,17875 \hfill \cr b = 0,08 \hfill \cr} \right. \to \left\{ \matrix{ \% Cu = 38,13\% \hfill \cr \% F{e_3}{O_4} = 61,87\% \hfill \cr} \right.\)
b.
\({Y_1}\left\{ \matrix{ CuS{O_4}:0,04 \hfill \cr FeS{O_4}:0,12 \hfill \cr {H_2}S{O_4}\,du:0,09 \hfill \cr} \right.\buildrel {{t^o}} \over \longrightarrow ran\,T\left\{ \matrix{ CuS{O_4}:0,04 \hfill \cr FeS{O_4}:0,12 \hfill \cr} \right. \to \left\{ \matrix{ \% CuS{O_4} = 25,97\% \hfill \cr FeS{O_4} = 74,03\% \hfill \cr} \right.\)
c. Ta dùng Ba(OH)2
\({Y_1}\left\{ \matrix{ CuS{O_4}:0,04 \hfill \cr FeS{O_4}:0,12 \hfill \cr {H_2}S{O_4}\,du:0,09 \hfill \cr} \right.\buildrel {Ba{{(OH)}_2}:\,0,2V\,mol} \over \longrightarrow \left\{ \matrix{ \downarrow \left\{ \matrix{ BaS{O_4}:0,25 \hfill \cr Cu{(OH)_2}:0,04 \hfill \cr Fe{(OH)_2}:0,12 \hfill \cr} \right. \hfill \cr {H_2}O \hfill \cr} \right.\buildrel {BTNT\,Ba} \over \longrightarrow Ba{(OH)_2}:\,0,25mol = > V = 1,25l\)