\(\begin{array}{l}
4)\\
a)\\
4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}\\
S{O_2} + B{r_2} + 2{H_2}O \to 2HBr + {H_2}S{O_4}\\
{H_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2HCl\\
b)\\
nBaS{O_4} = \dfrac{{2,33}}{{233}} = 0,01\,mol\\
= > nS{O_2} = n{H_2}S{O_4} = nBaS{O_4} = 0,01\,mol\\
VS{O_2} = 0,01 \times 22,4 = 0,224l\\
nFe{S_2} = \dfrac{{0,01}}{2} = 0,005\,mol\\
mFe{S_2} = 0,005 \times 120 = 0,6g\\
5)\\
a)\\
nNaOH = 0,25 \times 0,5 = 0,125\,mol\\
T = \dfrac{{nNaOH}}{{nS{O_2}}} = \dfrac{{0,125}}{{0,02}} = 6,25
\end{array}\)
T>2=> tạo muối $Na_2SO_3$
\(\begin{array}{l}
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
nN{a_2}S{O_3} = nS{O_2} = 0,02\,mol\\
mN{a_2}S{O_3} = 0,02 \times 126 = 2,52g\\
b)\\
nS{O_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
nNaOH = 0,5 \times 0,1 = 0,05\\
T = \dfrac{{nNaOH}}{{nS{O_2}}} = \dfrac{{0,05}}{{0,15}} = 0,33
\end{array}\)
T<1=> tạo muối $NaHSO_3$
\(\begin{array}{l}
NaOH + S{O_2} \to NaHS{O_3}\\
nNaHS{O_3} = nNaOH = 0,05\,mol\\
mNaHS{O_3} = 0,05 \times 104 = 5,2g
\end{array}\)
