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a, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Ta có: \(n_{CH_4}+n_{C_2H_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)
Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,2\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2.22,4}{6,72}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)
b, Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=0,7\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,7.22,4=15,68\left(l\right)\)
a. \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
a 2a a
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
b 3b 2b
b. \(n_{O_2}=\dfrac{25.88}{22.4}=1.155mol\)
n hỗn hợp khí \(=\dfrac{11.2}{22.4}=0.5mol\)
Ta có: \(\left\{{}\begin{matrix}a+b=0.5\\2a+3b=1.155\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.345\\b=0.155\end{matrix}\right.\)
\(\%V_{CH_4}=\dfrac{0.345\times22.4\times100}{11.2}=69\%\)
\(\%V_{C_2H_4}=100-69=31\%\)
c. \(V_{CO_2}=\left(a+2b\right)\times22.4=\left(0.345+2\times0.155\right)\times22.4=14.672l\)
\(n_{hhk}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(n_{O_2}=\dfrac{26,88}{22,4}=1,2\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\) ( mol ) \(\Rightarrow n_{hhk}=x+y=0,5\left(1\right)\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x 2x ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
y 3y ( mol )
\(\rightarrow n_{O_2}=2x+3y=1,2\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
\(\%V_{CH_4}=\dfrac{0,3}{0,5}.100=60\%\)
\(\%V_{C_2H_4}=100-60=40\%\)
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.
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x 2x ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
y 3y ( mol )
Ta có:
\(\left\{{}\begin{matrix}16x+28y=2,6\\2x+3y=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,075\\y=0,05\end{matrix}\right.\)
\(\Rightarrow m_{CH_4}=0,075.16=1,2g\)
\(\Rightarrow m_{C_2H_4}=0,05.28=1,4g\)
\(\%m_{CH_4}=\dfrac{1,2}{2,6}.100=46,15\%\)
\(\%m_{C_2H_4}=100\%-46,15\%=53,85\%\)
\(n_{CH_4}=22,4.0,075=1,68l\)
\(n_{C_2H_4}=0,05.22,4=1,12l\)
PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(1\right)\)
Theo PT: \(n_{O_2}=3n_{C_2H_4}+\dfrac{5}{2}n_{C_2H_2}=\dfrac{12,32}{22,4}=0,55\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow n_{C_2H_4}=n_{C_2H_2}=0,1\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,1.22,4}{4,48}.100\%=50\%\)
Giả sử các khí được đo ở điều kiện sao cho 1 mol khí chiếm thể tích 1 lít
Gọi số mol CH4, C2H2 là a, b (mol)
=> a + b = 56 (1)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
a--->2a
2C2H2 + 5O2 --to--> 4CO2 + 2H2O
b------>2,5b
=> 2a + 2,5b = 133,4 (2)
(1)(2) => a = 13,2 (mol); b = 42,8 (mol)
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{13,2}{56}.100\%=23,57\%\\\%V_{C_2H_2}=\dfrac{42,8}{56}.100\%=76,43\%\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\left(đk:0< a,b< 0,3\right)\)
\(n_{hh}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
PTHH:
CH4 + 2O2 --to--> CO2 + 2H2O
a------>a
C2H4 + 3O2 --to--> 2CO2 + 2H2O
b------->3b
=> Hệ pt \(\left\{{}\begin{matrix}a+b=0,3\\a+3b=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\left(TM\right)\)
\(\rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,1.22,4=2,24\left(l\right)\\V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)
Cảm ơn ông nhiều nha=)))