Đáp án:
a) 11,76l
b) 122,325g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
nMg = \dfrac{{3,6}}{{24}} = 0,15\,mol\\
nFe = \dfrac{{14}}{{56}} = 0,25\,mol\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
nS{O_2} = 0,15 + 0,25 \times 1,5 = 0,525\,mol\\
VS{O_2} = 0,525 \times 22,4 = 11,76l\\
nMgS{O_4} = nMg = 0,15\,mol\\
mMgS{O_4} = 0,15 \times 120 = 18g\\
nF{e_2}{(S{O_4})_3} = \dfrac{{0,25}}{2} = 0,125\,mol\\
mF{e_2}{(S{O_4})_3} = 0,125 \times 400 = 50g\\
b)\\
3BaC{l_2} + F{e_2}{(S{O_4})_3} \to 2FeC{l_3} + 3BaS{O_4}\\
MgS{O_4} + BaC{l_2} \to MgC{l_2} + BaS{O_4}\\
nBaS{O_4} = 0,125 \times 3 + 0,15 = 0,525\,mol\\
mBaS{O_4} = 0,525 \times 233 = 122,325g
\end{array}\)
