1)

$CH_4 +2 O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

Theo PTHH : 

$V_{O_2\ cần\ dùng} = 2V_{CH_4} = 24,79(lít)$
$V_{CO_2} = V_{CH_4} = 12,395(lít)$

2)

a)

$C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$
$V_{O_2} = 3V_{C_2H_4} = 14,874(lít)$

b) $V_{không\ khí} = V_{O_2} : 20\% = 14,874 : 20\% = 74,37(lít)$

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\Rightarrow n_{CO_2}=n_{CH_4}=0,5\left(mol\right);n_{O_2}=2.n_{CH_4}=2.0,5=1\left(mol\right)\\ V_{O_2\left(đktc\right)}=n_{O_2}.22,4=1.22,4=22,4\left(l\right)\\ V_{CO_2\left(đktc\right)}=n_{CO_2}.22,4=0,5.22,4=11,2\left(l\right)\)

\(n_{C_2H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{t^0}}2CO_2+H_2O\)

\(0.1........0.25...........0.2\)

\(V_{O_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(V_{CO_2}=0.2\cdot22.4=4.48\left(l\right)\)

n C2H2 = 22,4/22,4 = 1(mol)

$2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O$

Theo PTHH :

n O2 = 5/2 n C2H2 = 2,5(mol) => V O2 = 2,5.22,4 = 56(lít)

n CO2 = 2n C2H2 = 2(mol) => V CO2 = 2.22,4 = 44,8(lít)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

0.15     0.3      0.15

\(n_{CH_4}=\dfrac{3.36}{22.4}=0.15mol\)

\(V_{O_2}=0.3\times22.4=6.72l\)

\(V_{CO_2}=0.15\times22.4=3.36l\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(nO_2=3.0,25=0,75\left(mol\right)\)

\(VO_2=0,75.22,4=16,8\left(l\right)\)

\(nCO_2=2.0,25=0,5\left(mol\right)\)

\(VCO_2=0,5.224=11,2\left(l\right)\)

nCH4 = 2.24/22.4 = 0.1 (mol) 

CH4 + 2O2 -to-> CO2 + 2H2O 

0.1____0.2______0.1 

VO2 = 0.2*22.4 = 4.48 (l) 

VCO2 = 0.1*22.4=2.24 (l) 

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ V_{O_2} = 2V_{CH_4} = 2,24.2 = 4,48(lít)\\ V_{CO_2} = V_{CH_4} = 2,24(lít)\)

CH4 + 2O2 → CO2 + 2H2O

\(n_{CH4}\)=\(\dfrac{n_{CH4}}{22.4}=\)\(\dfrac{2.24}{22.4}=0.1\)mol

\(n_{O2}=\dfrac{0,1.2}{1}=0.2\)mol

\(V_{O2}=n_{O2}.22,4=0,2.22,4=4.48\)l

\(n_{CO2}=\dfrac{0,1.1}{1}0.1\)mol

\(V_{CO2}=n_{CO2}.22,4=0,1.22,4=2.24\)l

a, \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Ta có: \(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{O_2}=3n_{C_2H_4}=1,5\left(mol\right)\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

b, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=168\left(l\right)\)

PTHH: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PTHH: \(n_{O_2}=3.n_{C_2H_4}=3.0,5=1,5\left(mol\right)\)

=> \(V_{O_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\)

câu b oxi chiếm bao nhiêu của kk vậy bạn