PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=2,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=280\left(l\right)\)

\(2C_2H_2+5O_2-^{t^o}\rightarrow4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{1,3}{26}=0,05\left(mol\right)\\ n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,125\left(mol\right)\\ \Rightarrow V_{O_2}=2,8\left(l\right)\\ Vìtrongkhôngkhíchứa20\%O_2\\ \Rightarrow V_{kk}=\dfrac{2,8}{20\%}=14\left(l\right)\)

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

Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=1\left(mol\right)\Rightarrow V_{O_2}=1.22,4=22,4\left(l\right)\)

Đáp án: A

Câu 1:

\(a,PTHH:C_2H_4+5O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(n_{C_2H_4}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(Theo.PTHH:n_{O_2}=5.n_{C_2H_4}=5.0,25=1,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=n.22,4=1,25.22,4=28\left(l\right)\)

\(b,\Rightarrow V_{kk\left(đktc\right)}=5.V_{O_2\left(đktc\right)}=5.28=140\left(l\right)\)

B1:

\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)

Vì số mol tỉ lệ thuận với thể tích, đồng thời nhìn PTHH, ta sẽ được:

\(a,V_{O_2\left(đktc\right)}=3.V_{C_2H_4\left(đktc\right)}=3.5,6=16,8\left(l\right)\)

\(b,V_{kk}=5.V_{O_2\left(đktc\right)}=16,8.5=84\left(l\right)\)

B2:

Đặt số mol metan, etylen lần lượt là a,b (mol) (a,b>0)

\(n_{hh}=n_{CH_4}+n_{C_2H_4}=a+b=\dfrac{3,36}{22,4}=0,15\left(1\right)\)

PTHH: CH₄ +2 O₂ -t^o-> CO₂ +2 H₂O

C₂H₄ +3 O₂ -t^o-> 2CO₂ + 2H2O

\(n_{CO_2\left(tổng\right)}=a+2b=\dfrac{8,8}{44}=0,2\left(mol\right)\left(2\right)\)

(1), (2) =>a=0,1; b=0,05

Số mol tỉ lệ tương ứng với thể tích. Nên:

\(\%V_{CH_4}=\%n_{CH_4}=\dfrac{0,1}{0,15}.100\approx66,667\%\\ \Rightarrow\%V_{C_2H_4}\approx33,333\%\)

\(nC_2H_4=\frac{3,36}{22,4}=0,15mol\)

\(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\)

\(nO_2=3nC_2H_4=0,45mol\)

\(\rightarrow VO_2=0,45.24,79=11,1555l\)

PTHH: 2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

\(\dfrac{V_{O_2}}{V_{C_2H_2}}=\dfrac{n_{O_2}}{n_{C_2H_2}}=\dfrac{5}{2}\Rightarrow V_{O_2}=2,5.\dfrac{5}{2}=6,25\left(l\right)\)

=> \(V_{kk}=6,25.5=31,25\left(l\right)\)