- Gọi mol metan và etan là x, y ( mol )

\(x+y=n_{hh}=\dfrac{V}{22,4}=0,25\left(mol\right)\)

Lại có : \(x+2y=n_{CO_2}=\dfrac{V}{22,4}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\) ( mol )

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_4}=1,6\left(g\right)\\m_{C_2H_6}=4,5\left(g\right)\end{matrix}\right.\)

=> mhh = 6,1 ( g )

=> %mCH4 = ~ 26,22%

=> %mC2H6 = ~73,78%

Ta có : \(\%V_{CH4}=\dfrac{V}{Vhh}=40\%\)

=> %VC2H6 = 100 - %VCH4 = 60% .

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

\(2C_2H_6+5O_2\underrightarrow{t^o}4CO_2+6H_2O\)

Giả sử: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_6}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x+y=\dfrac{5,6}{22,4}=0,25\left(1\right)\)

Ta có: \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(\Sigma n_{CO_2}=n_{CH_4}+2n_{C_2H_6}\)

\(\Rightarrow x+2y=0,4\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,25}.100\%=40\%\\\%V_{C_2H_6}=60\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{0,1.16+0,15.30}.100\%\approx26,2\%\\\%m_{C_2H_6}\approx73,8\%\end{matrix}\right.\)

Bạn tham khảo nhé!

Gọi : 

\(n_{CH_4} = a(mol) ; n_{C_2H_6} = b(mol)\\ \Rightarrow a + b = \dfrac{5,6}{22,4} = 0,25(1)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_6 + \dfrac{7}{2}O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O\\ n_{CO_2} = a + 2b = \dfrac{8,96}{22,4} = 0,4(2)\)

Từ (1)(2) suy ra: a = 0,1 ; b = 0,15

Vậy :

\(\%V_{CH_4} = \dfrac{0,1.22,4}{5,6}.100\% = 40\%\\ \%V_{C_2H_6} = 100\% - 40\% = 60\%\\ \%m_{CH_4} = \dfrac{0,1.16}{0,1.16 +0,15.30}.100\% = 26,23\%\\ \%m_{C_2H_6} = 100\% - 26,23\% = 73,77\%\)

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

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

Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\left(1\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{56}{22,4}=2,5\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,5\left(mol\right)\\n_{C_2H_2}=1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,5.22,4}{33,6}.100\%\approx33,33\%\\\%V_{C_2H_2}\approx66,67\%\end{matrix}\right.\)

b, Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=3,5\left(mol\right)\Rightarrow m_{O_2}=3,5.32=112\left(g\right)\)

a.\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)

\(n_{C_2H_2Br_4}=\dfrac{6,72}{22,4}=0,3mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

 0,3                         0,3      ( mol )

\(\%C_2H_2=\dfrac{0,3}{0,6}.100=50\%\)

\(\%CH_4=100\%-50\%=50\%\)

b.

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 0,3       0,6                                       ( mol )

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

 0,3          0,75                                        ( mol )

\(V_{O_2}=\left(0,6+0,75\right).22,4=1,35.22,4=30,24l\)

\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)

\(CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(C_2H_2+\dfrac{5}{2}O_2\rightarrow2CO_2+H_2O\)

Từ hai pt trên:\(\Rightarrow\left\{{}\begin{matrix}x+y=0,15\\x+2y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\%V_{CH_4}=\dfrac{0,1}{0,1+0,05}\cdot100\%=66,67\%\)

\(\%V_{C_2H_2}=100\%-66,67\%=33,33\%\)

\(n_{CO_2}=\dfrac{V_{CO_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)

Gọi \(n_{CH_4}\) là x \(\Rightarrow V_{CH_4}=22,4x\)

      \(n_{C_2H_2}\) là y \(\Rightarrow V_{C_2H_2}=22,4y\)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

  x                              x             ( mol )

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

    y                              2y                    ( mol )

Ta có:

\(\left\{{}\begin{matrix}22,4x+22,4y=3,36\\x+2y=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\Rightarrow V_{CH_4}=22,4.0,1=2,24l\)

\(\Rightarrow V_{C_2H_2}=22,4.0,05=1,12l\)

\(\%V_{CH_4}=\dfrac{2,24}{3,36}.100=66,67\%\)

\(\%V_{C_2H_2}=100\%-66,67\%=33,33\%\)

nhh khí = 3,36/22,4 = 0,15 (mol)

Gọi nCH4 = a (mol); nC2H2 = b (mol)

a + b = 0,15 (1)

nCO2 = 4,48/22,4 = 0,2 (mol)

PTHH:

CH4 + 2O2 -> (t°) CO2 + 2H2O

a ---> 2a ---> a ---> 2a

2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O

b ---> 2,5b ---> 2b ---> b

a + 2b = 0,4 (2)

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

%VCH4 = 0,1/0,15 = 66,66%

%C2H2 = 100% - 66,66% = 33,34%

ko phải e làm đâu nha a , a tham khảo cho 

a)CH4 + 2O2 →→ CO2 + 2H2O (1)

2H2 + O2 →→ 2H2O (2)

b) Đặt nCH4 = a (mol) , nH2 = b (mol)

=> mCH4 = 16a(g) , mH2 = 2b(g)

mà n(CH4 + H2) = V/22,4 = 6,72/22,4 = 0,3(mol)

=> a + b = 0,3(mol) => a = 0,3 - b

Từ PT(1) => nO2 = 2nCH4 = 2a(mol)

Từ PT(2) => nO2 = 1/2 . nH2 = 1/2 . b (mol)

=>tổng nO2 = 2a + 1/2.b (mol)

=> mO2 = n .M = 32. (2a+ 1/2.b )= 64a + 16b(g)

Theo ĐLBTKL:mCH4 + mH2 + mO2 = mCO2 + mH2O(PT1,2)

=> 16a + 2b+ 64a +16b = 11,6

=> 80a + 18b = 11,6

=> 80. (0,3 - b ) + 18b = 11,6

=> b = 0,2(mol)

=> a = 0,3 - 0,2 = 0,1(mol)

=>mCH4 = 16a = 16. 0,1 = 1,6(g) , mH2 = 2b = 2 . 0,2 = 0,4(g)

=> VCH4 = n .22,4 = 0,1 . 22,4 =2,24(l)

VH2 = n . 22,4 = 0,2 . 22,4 = 4,48(l)

tự tính ... 

c) nO2 = 2a + 1/2 .b = 2 . 0,1 + 1/2 . 0,2 = 0,3(mol)

=> VO2 = 0,3 . 22,4 =6,72(l)

mà VO2 = 20% Vkk

=> Vkk = 6,72 : 20% =33,6(l)

\(n_{CH_4} = a\ mol ;n_{C_2H_6} = b\ mol\\ \Rightarrow a + b = \dfrac{3,36}{22,4}= 0,15(1)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_6 + \dfrac{7}{2}O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O\\ n_{CO_2} = a + 2b = \dfrac{4,48}{22,4} = 0,2(2)\\ (1)(2) \Rightarrow a = 0,1 ;b = 0,05\\ \Rightarrow \%V_{CH_4} = \dfrac{0,1}{0,15}.100\% = 66,67\%\\ \%V_{C_2H_6} = 100\% - 66,67\% = 33,33\%\)

\(n_{CH_4}=a\left(mol\right),n_{C_2H_6}=b\left(mol\right)\)

\(\Rightarrow a+b=0.15\left(mol\right)\left(1\right)\)

\(n_{CO_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(\Rightarrow a+2b=0.2\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.05\)

\(\%CH_4=\dfrac{0.1}{0.15}\cdot100\%=66.67\%\)

\(\%C_2H_6=33.33\%\)

Gọi số mol của metan là x, số mol của etan là y

nA = 0,150 mol = x+ y     (1)

(n_{CO_{2}}) = 0,20 mol = x + 2y      (2)

Từ (1) và (2)   => x = 0,100; y = 0,0500

=> %(V_{CH_{4}}) = 66,7% và %(V_{C_{2}H_{6}}) = 33,3%

CH₄ -> CO₂

C₂H₆ -> 2 CO₂

Gọi n_CH4 = x mol, n_C2H6 = y mol

x + y = 0,15 (1)

x + 2y = 0,2 (2)

Nên: x = 0,1 mol, y = 0,05 mol

Vậy: % V_CH4 = 66,67 % => %V_C2H6 = 33,33 %

Gọi: \(\left\{{}\begin{matrix}n_{CO}=x\left(mol\right)\\n_{H_2}=y\left(mol\right)\end{matrix}\right.\)

⇒ 28x + 2y = 11,8 (1)

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

\(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Ta có: \(n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{CO}+\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}x+\dfrac{1}{2}y\left(mol\right)\)

⇒ x + y = 0,7 (2)

Từ (1) và (2) ⇒ x = 0,4 (mol), y = 0,3 (mol)

a, \(\left\{{}\begin{matrix}\%m_{CO}=\dfrac{0,4.28}{11,8}.100\%\approx94,9\%\\\%m_{H_2}\approx5,1\%\end{matrix}\right.\)

b, Ở cùng điều kiện nhiệt độ và áp suất, % số mol cũng là % thể tích.

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,4}{0,7}.100\%\approx57,14\%\\\%V_{H_2}\approx42,86\%\end{matrix}\right.\)

Bạn tham khảo nhé!

PTHH: 

\(2CO+O_2\overset{t^o}{--->}2CO_2\left(1\right)\)

\(2H_2+O_2\overset{t^o}{--->}2H_2O\left(2\right)\)

Ta có: \(n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

Gọi x, y lần lượt là số mol của CO và H₂

a. Theo PT₍₁₎: \(n_{O_2}=\dfrac{1}{2}.n_{CO}=\dfrac{1}{2}x\left(mol\right)\)

Theo PT₍₂₎: \(n_{O_2}=\dfrac{1}{2}.n_{H_2}=\dfrac{1}{2}y\left(mol\right)\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{1}{2}y=0,35\) (*)

Theo đề, ta có: \(28x+2y=11,8\) (**)

Từ (*) và (**), ta có HPT:

\(\left\{{}\begin{matrix}\dfrac{1}{2}x+\dfrac{1}{2}y=0,35\\28x+2y=11,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,4\\y=0,3\end{matrix}\right.\)

\(\Rightarrow m_{H_2}=2.0,3=0,6\left(g\right)\)

\(\Rightarrow\%_{m_{H_2}}=\dfrac{0,6}{11,8}.100\%=5,08\%\)

\(\%_{m_{CO}}=100\%-5,08\%=94,92\%\)

b. \(\%_{V_{CO}}=\dfrac{0,4}{0,4+0,3}.100\%=57,1\%\)

\(\%_{V_{H_2}}=100\%-57,1\%=42,9\%\)

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

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

a, Giả sử: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x+y=\dfrac{22,4}{22,4}=1\left(mol\right)\left(1\right)\)

Ta có: \(n_{CO_2}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)

Theo PT: \(\Sigma n_{CO_2}=n_{CH_4}+2n_{C_2H_2}\)

\(\Rightarrow x+2y=1,6\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,4\left(mol\right)\\y=0,6\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,4}{1}.100\%=40\%\\\text{ }\%V_{C_2H_2}=60\%\end{matrix}\right.\)

b, Theo PT: \(\Sigma n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=2,3\left(mol\right)\)

\(\Rightarrow m_{O_2}=2,3.32=73,6\left(g\right)\)

c, PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{1,6}{0,8}=2M\)

Bạn tham khảo nhé!