\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1mol\)

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

0,1         0,2    0,1

\(V_{O_2}=0,2\cdot22,4=4,48l\)

\(V_{CO_2}=0,1\cdot22,4=2,24l\)

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

b, Ta có: \(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{O_2}=n_{H_2O}=2n_{CH_4}=0,5\left(mol\right)\\n_{CO_2}=n_{CH_4}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO_2}=0,25.44=11\left(g\right)\\m_{H_2O}=0,5.18=9\left(g\right)\end{matrix}\right.\)

c, Ta có: \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)

Mà: V_O2 = 1/5V_kk

\(\Rightarrow V_{kk}=11,2.5=56\left(l\right)\)

Bạn tham khảo nhé!

`a)` Gọi CTPT của hiđro cacbon là: `C_x H_y`

  `x=4/1=4`

`V_[H_2 O]=6.2-4.2=4(l)`

 `=>y=[4.2]/1=8`

`=>C_4 H_8`

`b)` 

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

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

b, Gọi: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_4}=y\left(mol\right)\end{matrix}\right.\) \(\Rightarrow x+y=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(1\right)\)

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=2x+3y=\dfrac{15,68}{22,4}=0,7\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=-0,1\\y=0,3\end{matrix}\right.\)

Đến đây thì ra số mol âm, bạn xem lại đề nhé.

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) \(n_{CH_4}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

b) Theo PTHH: \(n_{H_2O}=n_{O_2}=2n_{CH_4}=2.0,4=0,8\left(mol\right)\)

\(m_{H_2O}=n.M=0,8.18=14,4\left(g\right)\)

c) \(V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,8.22,4=17,92\left(l\right)\)

CH4+2O2-to>CO2+2H2O

0,1-----0,2

n CH4=0,1 mol

=>VO2=0,2.22,4=4,48l

a. \(n_{CH_4}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)

PTHH : CH₄ + 2O₂ ---t⁰---> CO₂ + 2H₂O

            0,2       0,4               0,2

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

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

c. \(V_{kk}=8,96.5=44,8\left(l\right)\)

\(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

          0,25-->0,5-------->0,25

=> \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)

\(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)