Đáp án:

 b) 13,44

c) $O_2$ dư 4,48l

4,48l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
b)\\
nKCl{O_3} = \frac{{49}}{{122,5}} = 0,4\,mol\\
 =  > n{O_2} = 0,6\,mol\\
V{O_2} = 0,6 \times 22,4 = 13,44l\\
c)\\
nC{H_4} = \frac{{4,48}}{{22,4}} = 0,2\,mol\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
\frac{{0,2}}{1} < \frac{{0,6}}{2}
\end{array}\)

=> $O_2$ dư

$nO_2$ dư=0,6-0,4=0,2mol

$VO_2$ dư=0,2x22,4=4,48l

n$CO_2$=n$CH_4$=0,2 mol

V$CO_2$=0,2x22,4=4,48l

$nKClO3=49/122,5=0,4mol$

$a/$

$2KClO3\overset{to}{\rightarrow}2KCl+3O2$

$b/$

theo pt :

$nO2=2/3.nKClO3=3/2.0,4=0,6mol$

$⇒V_{O2}=0,6.22,4=13,44l$

c/

$nCH4=4,48/22,4=0,2mol$

$pthh:$

$CH4+2O2\overset{to}{\rightarrow}CO2+2H2O$

⇒Sau phản ứng O2 dư

$⇒V_{O2 dư}=0,1.22,4=2,24l$

$V_{CO2}=0,2.22,4=4,48l$