Đáp án:

 Bạn tham khảo lời giải ở dưới nhé !

Giải thích các bước giải:

\(\begin{array}{l}
3.\\
c)\\
(1)2KCl{O_3} \to 2KCl + 3{O_2}\\
(2){O_2} + S \to S{O_2}\\
(3)S{O_2} + \dfrac{1}{2}{O_2} \to S{O_3}\\
(4)S{O_3} + {H_2}O \to {H_2}S{O_4}\\
(5)6{H_2}S{O_4} + 2Fe \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
(6)F{e_2}{(S{O_4})_3} + Fe \to 3FeS{O_4}
\end{array}\)

\(\begin{array}{l}
6.\\
a.\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{S{O_2}}} = 0,15mol\\
 \to {n_{Fe}} = \dfrac{2}{3}{n_{S{O_2}}} = 0,1mol\\
 \to {m_{Fe}} = 5,6g
\end{array}\)

\(\begin{array}{l}
b.\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,1mol\\
 \to {n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
 \to {V_{{H_2}}} = 2,24l
\end{array}\)

\(\begin{array}{l}
c.\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,1mol\\
 \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,2mol\\
 \to {V_{NaOH}} = 0,2l
\end{array}\)

\(\begin{array}{l}
8.\\
N{a_2}S{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + S{O_2} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = 0,1mol\\
 \to {n_{S{O_2}}} = {n_{N{a_2}S{O_3}}} = 0,1mol\\
 \to {V_{S{O_2}}} = 2,24l
\end{array}\)

Bài 3:

KClO3→KCl + 3/2 O2 (1) 

S + O2→SO2 (2)

SO2 + 1/2 O2→SO3 (3)

SO3 + H2O→H2SO4 (4)

6 H2SO4(đ/n) + 2Fe→Fe2(SO4)3+ 3 SO2 + 6H20 (5)

Fe2(SO4)3+Fe→3FeSO4 (6)

(Bạn thêm điều kiện giúp mình nhé)

Bài 6:

a, 6 H2SO4(đ/n) + 2Fe→Fe2(SO4)3+ 3 SO2 + 6H20

nS02= 3,36/22,4=0,15 (mol) => nFe= 0,15x2/3=0,1 (mol)

=> mFe=0,1x56=5,6 (g)

b, Fe + H2SO4→FeSO4+H2

nFe=nH2=5,6/56=0,1 (mol)  => VH2= 0,1x22,4=2,24 (l)

c, 2 NaOH+H2SO4→Na2SO4+ 2H20

nH2S04=0,1x1=0,1 (mol) => nNaOH=0,1x2=0,2 (mol)

=> VNaOH=0,2/1=0,2 (l)

Câu 8:

Na2SO3 + H2SO4 → Na2SO4 + SO2 + H2O

nNa2SO3=nSO2= 12,6/126=0,1 (mol) 

vSO2=0,1x22,4=2,24 (l)