Phản ứng cháy của benzen:

2 C 6 H 6 + 15 O 2 → 12 CO 2 + 6 H 2 O

n O 2  = 15/2. n C 6 H 6  = 0,75 mol

V O 2  = 0,75 x 22,4 = 16,8l

Nếu dùng không khí thì V kk  = 16,8/20 x 100 = 84l

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

Ta có: \(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

____0,2___0,4 (mol)

\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)

Bạn tham khảo nhé!

nCH4 = 4,48/22,4 = 0,2 (mol)

PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O

Mol: 0,2 ---> 0,4

VO2 = 0,4 . 22,4 = 8,96 (l)

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

Ta có: \(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

a, Theo PT: \(n_{O_2}=2n_{CH_4}=1\left(mol\right)\)

\(\Rightarrow V_{O_2}=1.22,4=22,4\left(l\right)\)

b, \(V_{kk}=\dfrac{22,4}{20\%}=112\left(l\right)\)

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:

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

\(a,\text{Đ}\text{ặt}:C_xH_y\left(x,y:nguy\text{ê}n,d\text{ươ}ng\right)\\ m_C=80\%.30=24\left(g\right)\Rightarrow x=n_C=\dfrac{24}{12}=2\\ m_H=20\%.30=6\left(g\right)\Rightarrow y=n_H=\dfrac{6}{1}=6\\ V\text{ới}:x=2;y=6\Rightarrow CTPT.X:C_2H_6\)

\(b,n_X=\dfrac{20,16}{22,4}=0,9\left(mol\right)\\ C_2H_6+\dfrac{7}{2}O_2\rightarrow2CO_2+3H_2O\\ n_{O_2}=\dfrac{7}{2}.0,9=3,15\left(mol\right)\\ n_{CO_2}=2.0,9=1,8\left(mol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=3,15.22,4=70,56\left(l\right)\\ V_{CO_2\left(\text{đ}ktc\right)}=1,8.22,4=40,32\left(l\right)\)

\(n_{C_2H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

0,1         0,25                                         ( mol )

\(V_{kk}=0,25.22,4.5=28l\)

\(n_{Mg}=\dfrac{13}{24}mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

\(\dfrac{13}{24}\)      \(\dfrac{13}{48}\)                          ( mol )

\(V_{O_2}=\dfrac{13}{48}.22,4=\dfrac{91}{15}l\)

2Mg+O2-to>2MgO

\(\dfrac{13}{24}\)---\(\dfrac{13}{48}\)

n Mg=\(\dfrac{13}{24}\)mol

=>VO2=\(\dfrac{13}{48}\).22,4=6,06l

nMg = 13 : 24 = 13/24 (mol) 
pthh : 2Mg + O2 -t--> 2MgO
          13/24->13/48 (mol) 
=> VO2 = 13/48 . 22,4 = 6,06 (l)