\(Đặt:n_{C_2H_2}=a\left(mol\right),n_{CH_4}=b\left(mol\right)\)

\(n_{hh}=a+b=0.15\left(mol\right)\left(1\right)\)

\(C_2H_2\rightarrow2CO_2\)

\(CH_4\rightarrow CO_2\)

\(n_{CO_2}=2a+b=0.2\left(mol\right)\left(2\right)\)

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

\(\%C_2H_2=\dfrac{0.05}{0.15}\cdot100\%=33.33\%\)

\(\%CH_4=66.67\%\)

\(2NaOH+CO_2\rightarrow Na_{_{ }2}CO_3+H_2O\)

\(0.4...............0.2............0.2\)

\(C_{M_{Na_2CO_3}}=\dfrac{0.2}{0.5}=0.4\left(M\right)\)

\(C_{M_{NaOH\left(dư\right)}}=\dfrac{0.5-0.4}{0.5}=0.2\left(M\right)\)

- 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)

\(m_{C_2H_2} = m_{tăng} = 5,2\ gam\\ \Rightarrow n_{C_2H_2} = \dfrac{5,2}{26} = 0,2(mol)\)

Vậy :

\(\%V_{C_2H_2} = \dfrac{0,2.22,4}{8,96}.100\% = 50\%\\ \%V_{CH_4} = 100\%-50\% = 50\%\)

b)

\(n_{CH_4} = n_{C_2H_2} = 0,2(mol)\)

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

0,2.........................0,2...................................(mol)

C₂H₂    +    \(\dfrac{5}{2}\)O₂ \(\xrightarrow{t^o}\)   2CO₂   +      H₂O

0,2................................0,4.................................(mol)

CO₂       +      Ca(OH)₂ → CaCO₃      +        H₂O

(0,2+0,4)............................(0,2+0,4)........................................(mol)

\(\Rightarrow m_{CaCO_3} =(0,2 + 0,4).100 = 60(gam)\)

Gọi số mol của metan và etan lần lượt là x và y (mol)

Phương trình phản ứng:

Vậy thành phần phần trăm về thể tích các khí trong hỗn hợp A là:

CH4+2O2-to>CO2+2H2O

x-----------------------------2x

C2H2+\(\dfrac{5}{2}\)O2-to>2CO2+H2O

y-----------------------------------y

=>\(\left\{{}\begin{matrix}x+y=\dfrac{3,36}{22,4}\\2x+y=\dfrac{4,5}{18}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

=>%VCH4=\(\dfrac{0,1.22,4}{3,36}\).100=66,67%

=>%VC2H2=100-66,67%=33,33%

b)

C2H2+2Br2->C2H2Br4

0,05-----0,1 mol

=>m Br2=0,1.160=16g

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

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

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

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

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

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

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

\(\%C_2H_6=33.33\%\)

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 %

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!

a) 

\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

             0,05<-0,05

=> \(n_{CH_4}=\dfrac{3,36}{22,4}-0,05=0,1\left(mol\right)\)

\(\%m_{CH_4}=\dfrac{0,1.16}{0,1.16+0,05.28}.100\%=53,33\%\)

\(\%m_{C_2H_4}=\dfrac{0,05.28}{0,1.16+0,05.28}.100\%=46,67\%\)

b)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,1-->0,2

            C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

           0,05--->0,15

=> \(V_{O_2}=\left(0,2+0,15\right).22,4=7,84\left(l\right)\)