Đáp án:

 Bài 3: a) $\dfrac{{158}}{{101}}$

b) $\dfrac{{606}}{{245}}$

Bài 4: 

a) 17 g

b) $m = 1,28g$; ${V_{C{O_2}}} = 2,1952(l)$; ${V_{S{O_2}}} = 0,03584(l)$

Giải thích các bước giải:

 Bài 3: 

a) $2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}$ (1)

$2KN{O_3} \to 2KN{O_2} + {O_2}$ (2)

Ta có: ${n_{{O_2}(1)}} = \dfrac{1}{2}{n_{KMn{O_4}}};{n_{{O_2}(2)}} = \dfrac{1}{2}{n_{KN{O_3}}}$

Mà ${n_{{O_2}(1)}} = {n_{{O_2}(2)}} \Rightarrow {n_{KMn{O_4}}} = {n_{KN{O_3}}}$

$ \Rightarrow \dfrac{x}{y} = \dfrac{{{m_{KMn{O_4}}}}}{{{m_{KN{O_3}}}}} = \dfrac{{158}}{{101}}$

b) $2KCl{O_3} \to 2KCl + 3{O_2}$ (1)

$2KN{O_3} \to 2KN{O_2} + {O_2}$ (2)

Ta có: ${n_{{O_2}(1)}} = \dfrac{3}{2}{n_{KCl{O_3}}} = \dfrac{3}{2}.\dfrac{m}{{122,5}} = \dfrac{{3m}}{{245}}$

${n_{{O_2}(2)}} = \dfrac{1}{2}{n_{KN{O_3}}} = \dfrac{1}{2}.\frac{m}{{101}} = \dfrac{m}{{202}}$

$ \Rightarrow \dfrac{x}{y} = \dfrac{{{n_{{O_2}(1)}}}}{{{n_{{O_2}(2)}}}} = \dfrac{{\dfrac{{3m}}{{245}}}}{{\dfrac{m}{{202}}}} = \dfrac{{606}}{{245}}$

Bài 4: 

a) $2KN{O_3} \to 2KN{O_2} + {O_2}$

${n_{KN{O_3}}} = \dfrac{{20,2}}{{101}} = 0,2mol \Rightarrow {n_{KN{O_2}}} = {n_{KN{O_3}}} = 0,2mol$

$ \Rightarrow {m_{KN{O_2}}} = 0,2.85 = 17g$

${n_{{O_2}}} = \dfrac{1}{2}{n_{KN{O_3}}} = 0,1mol$

b) 

$\begin{gathered}
  C + {O_2} \to C{O_2}{\text{     (1)}} \hfill \\
  S + {O_2} \to S{O_2}{\text{      (2)}} \hfill \\ 
\end{gathered} $

Ta có: ${m_C} = 0,92m \Rightarrow {n_C} = \dfrac{{23m}}{{300}};{m_S} = 0,04m \Rightarrow {n_S} = 1,{25.10^{ - 3}}m$

Theo pt (1) và (2): ${n_{{O_2}}} = {n_C} + {n_S} = \dfrac{{23m}}{{300}} + 1,{25.10^{ - 3}}m = 0,1 \Rightarrow m = 1,28g$

${n_{C{O_2}}} = {n_C} = \dfrac{{23.1,28}}{{300}} = 0,098mol \Rightarrow {V_{C{O_2}}} = 2,1952(l)$

${n_{S{O_2}}} = {n_S} = 1,{25.10^{ - 3}}.1,28 = 1,{6.10^{ - 3}}mol \Rightarrow {V_{S{O_2}}} = 0,03584(l)$