\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
- PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)
0,3<-0,15
=> VCO = 0,3.22,4 = 6,72 (l)
- PTHH: \(2C_2H_6+7O_2\underrightarrow{t^o}4CO_2+6H_2O\)
\(\dfrac{0,3}{7}\)<--0,15
=> \(V_{C_2H_6}=\dfrac{0,3}{7}.22,4=0,96\left(l\right)\)
- PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
0,075<--0,15
=> \(V_{CH_4}=0,075.22,4=1,68\left(l\right)\)
$2CO + O_2 \xrightarrow{t^o} 2CO_2$
$2C_2H_6 + 7O_2 \xrightarrow{t^o} 4CO_2 + 6H_2O$
$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
Theo PTHH :
$V_{CO} = 2V_{O_2} = 3,36.2 = 6,72(lít)$
$V_{C_2H_6} = \dfrac{2}{7}V_{O_2} = 0,96(lít)$
$V_{CH_4} = \dfrac{1}{2}V_{O_2} = 1,68(lít)$
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH:
\(2CO+O_2\xrightarrow[]{t^o}2CO_2\)
0,3<---0,15
\(2C_2H_6+7O_2\xrightarrow[]{t^o}4CO_2+6H_2O\)
\(\dfrac{3}{70}\)<----0,15
\(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,075<-0,15
=> \(\left\{{}\begin{matrix}V_{CO}=0,3.22,4=6,72\left(l\right)\\V_{C_2H_6}=\dfrac{3}{70}.22,4=0,96\left(l\right)\\V_{CH_4}=0,075.22,4=1,68\left(l\right)\end{matrix}\right.\)