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a) C2H4 + Br2 ---> C2H4Br2
Khí thoát ra là CH4 (metan) ---> mCH4 = 0,2.16 = 3,2g ---> mC2H4 = 6 - 3,2 = 2,8g. (0,1 mol).
---> %CH4 = 0,2/0,3 = 66,7%; %C2H4 = 33,3%.
b) nCO2 = nCH4 + 2nC2H4 = 0,4 mol.
CO2 + Ca(OH)2 ---> CaCO3 + H2O
CO2 + CaCO3 + H2O ---> Ca(HCO3)2
---> nCaCO3 = 0,2 mol ---> m = 0,2.100 = 20g.
Sửa : 29,25 \(\to\) 29,55
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = \dfrac{8}{160} = 0,05(mol)\\ \Rightarrow m_{C_2H_4} = 0,05.28 = 1,4(gam)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 +3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ CO_2 + Ba(OH)_2 \to BaCO_3 + H_2O\\ n_{CO_2} = n_{CH_4} + 2n_{C_2H_4} = n_{CH_4} + 0,05.2 = n_{BaCO_3} = \dfrac{29,55}{197}=0,15(mol) \\ \Rightarrow n_{CH_4} = 0,05(mol)\\ \Rightarrow m_{CH_4} = 0,05.16 = 0,8(gam)\)
Đầu tiên, không có nước Br chỉ có nước Br2 em nhé!
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nBr2= 8/160=0,05(mol)
PTHH: C2H4 + Br2 -> C2H4Br2
nC2H4=nBr2=0,05(mol) => mC2H4=0,05.28=1,4(g)
- Khí bay ra là khí CH4.
CH4 + 2 O2 -to-> CO2 + 2 H2O
CO2 + Ba(OH)2 -> BaCO3 + H2O
nBaCO3=29,25/197= 117/ 788 (mol ) (Số xấu quá em ơi)
=> nCH4=nCO2=nBaCO3= 117/788(mol)
=> mCH4=16. 117/788= 468/197(g)
\(n_{hh}=6,72:22,4=0,3mol\\ C_2H_2+2Br_2->C_2H_2Br_4\\ C_2H_4+Br_2->C_2H_2Br_2\\ n_{Br_2}=0,4mol\\ n_{C_2H_2}=a;n_{C_2H_4}=b\\ a+b=0,3\\ 2a+b=0,4\\ a=0,2;b=0,1\\ \%V_{C_2H_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\ \%V_{C_2H_4}=33,33\%\)
1/2 hỗn hợp có 0,1 mol C2H2 và 0,05mol C2H4
\(BT.C:n_{CO_2}=2n_{C_2H_2}+2n_{C_2H_4}=0,3mol\\ n_{CaCO_3}=n_{CO_2}=0,3\\ m_{KT}=0,3.100=30g\)
\(a,n_{hh\left(CH_4,C_2H_4,C_2H_2\right)}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{hh\left(C_2H_4,C_2H_2\right)}=0,4-0,1=0,3\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a+b=0,3\\28a+26b=8,1\end{matrix}\right.\Leftrightarrow a=b=0,15\left(mol\right)\)
PTHH:
\(CH\equiv CH+2Br-Br\rightarrow CHBr_2-CHBr_2\)
\(CH_2=CH_2+Br-Br\rightarrow CH_2Br-CH_2Br\)
\(b,\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,4}.100\%=25\%\\\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,15}{0,4}.100\%=37,5\%\end{matrix}\right.\)
c, PTHH:
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\\ C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\\ \rightarrow n_{BaCO_3}=n_{CO_2}=0,1+0,15.0,15.2=0,7\left(mol\right)\\ m_{BaCO_3}=0,7.197=137,9\left(g\right)\)
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)\(\%V_{metan}=\dfrac{0,2}{0,25}\cdot100\%=80\%\)
\(\%V_{etilen}=100\%-80\%=20\%\)
Trong hai khí trên, chỉ có khí etilen tác dụng với dung dịch nước brom, khối lượng bình tăng bằng khối lượng khí etilen tham gia phản ứng, suy ra số mol của khí etilen là 2,8/28=0,1 (mol).
Khí đem đốt cháy là khí metan, số mol khí metan bằng số mol khí CO2 sinh ra (bảo toàn nguyên tố cacbon) và bằng 11/44=0,25 (mol).
a. Giá trị V=(0,25+0,1).22,4=7,84 (lít).
b. Phần trăm thể tích mỗi khí:
%Vmetan=0,25/0,35.100%\(\approx\)71,43%, suy ra %Vetilen\(\approx\)100%-71,43%\(\approx\)28,57%.
a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,05\left(mol\right)\)
\(\Rightarrow V_{C_2H_2}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow V_{CH_4}=3,36-1,12=2,24\left(l\right)\)
b, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,325\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,325.22,4=7,28\left(l\right)\Rightarrow V_{kk}=5V_{O_2}=36,4\left(l\right)\)
Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=0,2\left(mol\right)\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
\(\Rightarrow n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CaCO_3}=0,2.100=20\left(g\right)\)
a)
PTHH: C2H2+2Br2 --> C2H2Br4
b) \(n_{C_2H_2}=\dfrac{36}{26}=\dfrac{18}{13}\left(mol\right)\)
=> \(V_{C_2H_2}=\dfrac{18}{13}.22,4=\dfrac{2016}{65}\left(l\right)\)
\(n_{CH_4}=\dfrac{42-36}{16}=0,375\left(mol\right)\)
=> \(V_{CH_4}=0,375.22,4=8,4\left(l\right)\)
c) \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{\dfrac{2016}{65}}{\dfrac{2016}{65}+8,4}.100\%=78,69\%\\\%V_{CH_4}=\dfrac{8,4}{\dfrac{2016}{65}+8,4}.100\%=21,31\%\end{matrix}\right.\)
a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có: \(n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\)
Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,2\left(mol\right)\)
Mà: \(n_X=\dfrac{10,08}{22,4}=0,45\left(mol\right)\Rightarrow n_{CH_4}=0,25\left(mol\right)\)
\(\Rightarrow V_{CH_4}=0,25.22,4=5,6\left(l\right)\)
c, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=0,65\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,65\left(mol\right)\Rightarrow m_{CaCO_3}=0,65.100=65\left(g\right)\)