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C2H4+Br2->C2H4Br2
0,05----0,05
n Br2=\(\dfrac{8}{160}\)=0,05 mol
=>%VC2H4=\(\dfrac{0,05.22,4}{5,6}.100=20\%\)
=>%VCH4=80%
c)CH4+2O2-to>CO2+2H2O
1.10-3----2.10-3 mol
C2H4+3O2-to>2CO2+2H2O
2,5.10-4-7,5.10-4 mol
n hh=\(\dfrac{0,028}{22,4}\)=1,25.10-3 mol
=>n C2H4=2,5.10-4 mol
=>n CH4=1.10-3 mol
=>VO2=(2.10-3+7,5.10-4).22,4=0,0616l
\(n_{Br_2}=\dfrac{8}{160}=0,05mol\)
\(\Rightarrow n_{etilen}=n_{Br_2}=0,05mol\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(\Rightarrow n_{metan}=n_{hh}-n_{etilen}=0,25-0,05=0,2mol\)
a)\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b)\(\%V_{metan}=\dfrac{0,2}{0,25}\cdot100\%=80\%\)
\(\%V_{etilen}=100\%-80\%=20\%\)
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 \(\left\{{}\begin{matrix}a+b=0,3\\a+3b=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\left(TM\right)\)
\(\rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,1.22,4=2,24\left(l\right)\\V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)
a) PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
b) Đặt \(n_{CH_4}=x\left(mol\right);n_{C_2H_4}=y\left(mol\right)\). Khi đó \(22,4x+22,4y=4,48\) \(\Leftrightarrow x+y=0,2\)
Từ PTHH \(\Rightarrow n_{O_2\left(1\right)}=2x\left(mol\right)\)\(;n_{O_2\left(2\right)}=3y\left(mol\right)\). Khi đó \(2x.22,4+3y.22,4=11,2\) \(\Leftrightarrow2x+3y=0,5\)
Vậy ta có \(\left\{{}\begin{matrix}x+y=0,2\\2x+3y=0,5\end{matrix}\right.\Leftrightarrow x=y=0,1\left(mol\right)\)
\(\Rightarrow\%V_{CH_4}=\%V_{C_2H_4}=50\%\)
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. \(V_{CO_2}=\left(a+2b\right)\times22.4=\left(0.345+2\times0.155\right)\times22.4=14.672l\)
a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)
\(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,25\left(mol\right)\\n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,25.22,4}{6,72}.100\%\approx83,33\%\\\%V_{C_2H_2}\approx16,67\%\end{matrix}\right.\)
Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,625\left(mol\right)\Rightarrow m_{O_2}=0,625.32=20\left(g\right)\)
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: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=0,7\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,7.22,4=15,68\left(l\right)\)
\(a)n_{Br_2} =\dfrac{16}{160} = 0,1(mol) C_2H_2 +2Br_2 \to C_2H_2Br_4\\ n_{C_2H_2} = \dfrac{1}{2}n_{Br_2} = 0,05(mol)\\ V_{C_2H_2} = 0,05.22,4 = 1,12(lít)\\ V_{CH_4} = 6,72 -1,12 = 5,6(lít)\\ b)\%V_{C_2H_2} = \dfrac{1,12}{6,72}.100\% = 16,67\%\\ \%V_{CH_4} = 100\%-16,67\% = 83,33\%\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\)
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:
\(\left\{{}\begin{matrix}16x+28y=2,6\\2x+3y=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,075\\y=0,05\end{matrix}\right.\)
\(\Rightarrow m_{CH_4}=0,075.16=1,2g\)
\(\Rightarrow m_{C_2H_4}=0,05.28=1,4g\)
\(\%m_{CH_4}=\dfrac{1,2}{2,6}.100=46,15\%\)
\(\%m_{C_2H_4}=100\%-46,15\%=53,85\%\)
\(n_{CH_4}=22,4.0,075=1,68l\)
\(n_{C_2H_4}=0,05.22,4=1,12l\)