Đáp án:
\(\begin{array}{l}
13,C\\
14,B\\
15,C\\
16,B\\
17,B\\
18,A\\
19,B\\
20,C\\
21,A\\
22,D\\
23,A\\
24,D\\
25,B\\
26,B\\
27,C\\
28,B\\
29,\\
a,S\\
b,S\\
c,S\\
d,Đ
\end{array}\)
Giải thích các bước giải:
21,
\(\begin{array}{l}
2KCl{O_3} \to 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = 1mol\\
\to {n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 1,5mol\\
\to {V_{{O_2}}} = 33,6l
\end{array}\)
22,
\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{{O_2}}} = 0,1mol\\
\to {n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,2mol\\
\to {m_{KMn{O_4}}} = 31,6g
\end{array}\)
28,
\(\begin{array}{l}
2KCl{O_3} \to 2KCl + 3{O_2}\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
\to {n_{{O_2}}}(KCl{O_3}) = \dfrac{3}{2}{n_{KCl{O_3}}} = 1,5amol\\
\to {n_{{O_2}}}(KMn{O_4}) = \dfrac{1}{2}{n_{KMn{O_4}}} = 0,5amol\\
\to {n_{{O_2}}}(KCl{O_3}) > {n_{{O_2}}}(KMn{O_4})\\
\to \dfrac{{{V_{{O_2}}}(KCl{O_3})}}{{{V_{{O_2}}}(KMn{O_4})}} = \dfrac{{1,5a}}{{0,5a}} = 3:1
\end{array}\)
