Đáp án:
2) 98g
3) 58,6g
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
0,1\,\,\,\,\,\,\,0,1\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
0,1\,\,\,\,\,\,\,\,0,1\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,2 \times 98}}{{20\% }} = 98g\\
3)\\
FeS{O_4} + Ba{(OH)_2} \to BaS{O_4} + Fe{(OH)_2}\\
0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,0,1\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \to 4Fe{(OH)_3}\\
0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,05\\
MgS{O_4} + Ba{(OH)_2} \to BaS{O_4} + Mg{(OH)_2}\\
0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\\
Mg{(OH)_2} \to MgO + {H_2}O\\
0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\\
m = mMgO + mF{e_2}{O_3} + mBaS{O_4} = 58,6g
\end{array}\)
