a) \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

          0,05-->0,1------->0,05

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

            0,125<--0,3125<----0,25

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,05+0,125}.100\%=28,57\%\\\%V_{C_2H_2}=\dfrac{0,125}{0,05+0,125}.100\%=71,43\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,05.16}{0,05.16+0,125.26}.100\%=19,753\%\\\%m_{C_2H_2}=\dfrac{0,125.26}{0,05.16+0,125.26}.100\%=80,247\%\end{matrix}\right.\)

b) \(n_{O_2}=0,1+0,3125=0,4125\left(mol\right)\)

=> \(V_{O_2}=0,4125.22,4=9,24\left(l\right)\)

=> V_kk = 9,24.5 = 46,2 (l)

nhh khí = 5,6/22,4 = 0,25 (mol)

nBr2 = 16/160 = 0,1 (mol)

PTHH: C2H2 + 2Br2 -> C2H2Br4

Mol: 0,05 <--- 0,1

nCH4 = 0,25 - 0,05 = 0,2 (mol)

%VC2H2 = 0,05/0,25 = 20%

%VCH4 = 100% - 20% = 80%

PTHH: 

2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O

0,05 ---> 0,125 ---> 0,1

CH4 + 2O2 -> (t°) CO2 + 2H2O

0,2 ---> 0,4 ---> 0,2

nCO2 = 0,2 + 0,1 = 0,3 (mol)

PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2O

nCaCO3 = 0,3 (mol)

mCaCO3 = 0,3 . 100 = 30 (g)

Giả sử các khí đo ở điều kiện sao cho 1 mol khí chiếm thể tích V lít

Gọi số mol CH₄, C₂H₂ là a, b (mol)

=> \(a+b=\dfrac{0,05}{V}\) (1)

\(n_{O_2}=\dfrac{0,11}{V}\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

               a-->2a---------->a

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

                b--->2,5b-------->2b

=> \(2a+2,5b=\dfrac{0,11}{V}\) (2)

(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{0,03}{V}\left(mol\right)\\b=\dfrac{0,02}{V}\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{\dfrac{0,03}{V}.V}{0,05}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{\dfrac{0,02}{V}.V}{0,05}.100\%=40\%\end{matrix}\right.\)

b) \(n_{CO_2}=a+2b=\dfrac{0,07}{V}\left(mol\right)\)

=> \(V_{CO_2}=\dfrac{0,07}{V}.V=0,07\left(l\right)\)

Tham khảo:

a)

{VC2H2=x(ml)

VC2H4=y(ml)

⇒ x + y = 50(1)

C2H2+5/2O2(to)→2CO2+H2O

C2H4+3O2(to)→2CO2+2H2O

Theo PTHH : 2,5x + 3y = 140(2)

Từ (1)(2) suy ra: x = 20 ; y = 30

Vậy :

%VC2H2=2050.100%=40%%

VC2H4=100%−40%=60%

b)VCO2=2VC2H2+2VC2H4=2.50=100(ml)

a)

{VC2H2=x(ml)

VC2H4=y(ml)

⇒ x + y = 50(1)

C2H2+5/2O2(to)→2CO2+H2O

C2H4+3O2(to)→2CO2+2H2O

Theo PTHH : 2,5x + 3y = 140(2)

Từ (1)(2) suy ra: x = 20 ; y = 30

Vậy :

%VC2H2=2050.100%=40%%

VC2H4=100%−40%=60%

b)VCO2=2VC2H2+2VC2H4=2.50=100(ml)

a, $CH_4+2O_2\rightarrow CO_2+2H_2O$

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

b, Gọi thể tích CH4 và C2H2 lần lượt là a;b

Ta có: $a+b=14;2a+2,5b=33,6\Rightarrow a=2,8;b=11,2$

$\Rightarrow \%V_{CH_4}=20\%;\%V_{C_2H_2}=80\%$

c, $CO_2+Ca(OH)_2\rightarrow CaCO_3+H_2O$

Ta có: $n_{CaCO_3}=n_{CO_2}=\frac{25,2.10^{-3}}{22,4}=1,125.10^{-3}$

$\Rightarrow m=0,11259(g)$

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

b, \(n_{CH_4}=\dfrac{9,6}{16}=0,6\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}=0,6\left(mol\right)\Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\)

c, \(n_{O_2}=2n_{CH_4}=1,2\left(mol\right)\Rightarrow V_{O_2}=1,2.22,4=26,88\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=134,4\left(l\right)\)

\(n_{hhkhí}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ Gọi\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\)

PTHH: 

CH₄ + 2O₂ \(\underrightarrow{t^o}\) CO₂ + 2H₂O

   a       2a         a

2C₂H₂ + 5O₂ \(\underrightarrow{t^o}\) 4CO₂ + 2H₂O

b              2,5b      2b

Hệ phương trình: \(\left\{{}\begin{matrix}a+b=0,125\\2a+2,5b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,025\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)

\(\%V_{C_2H_2}=\dfrac{0,1}{0,125}=80\%\\ \%_{CH_4}=100\%-80\%=20\%\)

n_CO2 = 2.0,025 + 2.0,1 = 0,25 (mol)

PTHH: Ca(OH)₂ + CO₂ -> CaCO₃ + H₂O

                                0,25      0,25

=> m_CaCO3 = 0,25.100 = 25 (g)

Tính % thể tích các khí :

% V C 2 H 2  = 0,448/0,896 x 100% = 50%

% V CH 4  = % V C 2 H 6  = 25%

\(n_{CH_4}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(0.05.....0.1.......0.05\)

\(V_{O_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(V_{CO_2}=0.05\cdot22.4=1.12\left(l\right)\)