Đáp án:
1,D
2,D
Giải thích các bước giải:
1,
\(\begin{array}{l}
F{e_3}{O_4} + 4{H_2}S{O_4} \to FeS{O_4} + F{e_2}{(S{O_4})_3} + 4{H_2}O\\
{n_{F{e_3}{O_4}}} = 0,075mol\\
\to {n_{{H_2}S{O_4}}} = 0,3mol \to {m_{{H_2}S{O_4}}} = 29,4g\\
\to {m_{{H_2}S{O_4}}}dd = \dfrac{{29,4 \times 100}}{{19,6}} = 150g
\end{array}\)
2,
\(\begin{array}{l}
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
F{e_2}{(S{O_4})_3} + 3Ba{(OH)_2} \to 3BaS{O_4} + 2Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
{n_{S{O_2}}} = 0,15mol\\
{n_{Fe}} = \dfrac{2}{3}{n_{S{O_2}}} = 0,1mol\\
\to {m_{Fe}} = 5,6g \to {m_{F{e_2}{O_3}}} = 16g \to {n_{F{e_2}{O_3}}} = 0,1mol
\end{array}\)
Nung Z thu được chất rắn gồm: \(BaS{O_4}\) và \(F{e_2}{O_3}\)
\(\begin{array}{l}
\to {n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Fe}} + {n_{F{e_2}{O_3}}} = 0,15mol\\
\to {n_{BaS{O_4}}} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,45mol\\
\to {n_{Fe{{(OH)}_2}}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,3mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe{{(OH)}_2}}} = 0,15mol\\
\to m = {m_{BaS{O_4}}} + {m_{F{e_2}{O_3}}} = 128,85g
\end{array}\)