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a) mtăng = mC2H4 = 7 (g)
=> \(n_{C_2H_4}=\dfrac{7}{28}=0,25\left(mol\right)\)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,25.24,79}{9,916}.100\%=62,5\%\\\%V_{CH_4}=100\%-62,5\%=37,5\%\end{matrix}\right.\)
b)
PTHH: C2H4 + Br2 ---> C2H4Br2
0,25-->0,25
=> \(C_{M\left(dd.Br_2\right)}=\dfrac{0,25}{0,25}=1M\)
c)
\(n_{CH_4}=\dfrac{9,916}{24,79}-0,25=0,15\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,15-->0,3
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,25-->0,75
=> VO2(đkc) = (0,3 + 0,75).24,79 = 26,0295 (l)
a)
nBr2 = 0,2.0,2 = 0,04 (mol)
nCaCO3 = \(\dfrac{10}{100}=0,1\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,04<--0,04---->0,04
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,04--------------->0,08
CH4 + 2O2 --to--> CO2 + 2H2O
0,02<-------------0,02
Ca(OH)2 + CO2 --> CaCO3 + H2O
0,1<------0,1
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,02}{0,02+0,04}.100\%=33,33\%\\\%V_{C_2H_4}=\dfrac{0,04}{0,02+0,04}.100\%=66,67\%\end{matrix}\right.\)
b) mC2H4Br2 = 0,04.188 = 7,52 (g)
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(m_{tăng}=m_{Br_2}=m_{C_2H_2}=2,6g\)
\(\Rightarrow n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,1 0,1
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\)
0,1 0,25 0,2
\(\Rightarrow n_{CO_2\left(CH_4\right)}=0,4-0,2=0,2mol\)
\(\Rightarrow n_{CH_4}=0,2mol\Rightarrow n_{O_2}=0,4mol\)
a)\(\%V_{CH_4}=\dfrac{0,2}{0,4}\cdot100\%=50\%\)
\(\%V_{C_2H_2}=100\%-50\%=50\%\)
b)\(\Sigma n_{O_2}=0,4+0,25=0,65mol\)
\(\Rightarrow V_{O_2}=0,65\cdot22,4=14,56l\)
\(\Rightarrow V_{kk}=14,56\cdot5=72,8l\)
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: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,625\left(mol\right)\Rightarrow m_{O_2}=0,625.32=20\left(g\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
Ta có: \(n_{CO_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)=n_{CH_4}\)
Đặt \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow a+b=\dfrac{5,04}{22,4}-0,075=0,15\) (1)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Theo PTHH: \(28a+26b=4,1\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{C_2H_4}=0,1\left(mol\right)\\b=n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)
Mặt khác: \(n_{hh}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,075}{0,225}\cdot100\%\approx33,33\%\\\%V_{C_2H_4}=\dfrac{0,1}{0,225}\cdot100\%\approx44,44\%\\\%V_{C_2H_2}=22,23\%\end{matrix}\right.\)
\(a,Gọi\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{C_2H_2}=c\left(mol\right)\end{matrix}\right.\\ n_{hhkhí}=0,4\left(mol\right)\\ n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:a\rightarrow2a\rightarrow a\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:b\rightarrow3b\rightarrow2b\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:c\rightarrow2,5c\rightarrow2c\\ Hệ.pt\left\{{}\begin{matrix}a+b+c=0,4\\b+2c=0,4\\a+2b+2c=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\)
\(\%V_{CH_4}=\%V_{C_2H_2}=\dfrac{0,1}{0,4}=25\%\\ \%V_{C_2H_4}=\dfrac{0,2}{0,4}=50\%\)
\(m_{CH_4}=0,1.16=1,6\left(g\right)\\ m_{C_2H_4}=28.0,2=5,6\left(g\right)\\ m_{C_2H_2}=0,1.26=2,6\left(g\right)\\ \%m_{CH_4}=\dfrac{1,6}{1,6+5,6+2,6}=16,32\%\\ \%m_{C_2H_4}=\dfrac{5,6}{1,6+5,6+2,6}=57,14\%\\ \%m_{C_2H_2}=100\%-16,32\%-57,14\%=26,54\%\)
\(b,PTHH:C_2H_5OH\rightarrow C_2H_4+H_2O\\ Mol:0,2\leftarrow0,2\\ m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)
Dài quá!!!
a) Khi cho metan và axetilen qua dung dịch brom thì metan không phản ứng với brom nên thoát ra khỏi bình còn axetilen phản ứng với dung dịch brom.
=> 20,16 lít khí thoát là metan CH4
=> V axetilen = 40,32 - 20,16 = 20,16 lít
<=> %V CH4 = %V C2H2 = 50%
b)
nCH4 = nC2H2 = \(\dfrac{20,16}{22,4}\)= 0,9 lít
CH4 + 2O2 → CO2 + 2H2O
C2H2 + \(\dfrac{5}{2}\)O2 → 2CO2 + H2O
Theo tỉ lệ phản ứng cháy => nO2 cần để đốt cháy hết hỗn hợp khí = 2nCH4+\(\dfrac{5}{2}\)nC2H2= 4,05 mol.
=> V O2 cần dùng = 4,05.22,4 = 90,72 lít
a) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{6,4}{160}=0,04\left(mol\right);n_{hh}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,04<--0,04
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,04}{0,6}.100\%=6,67\%\\\%V_{CH_4}=100\%-6,67\%=93,33\%\end{matrix}\right.\)
b) \(n_{CH_4}=0,6-0,04=0,56\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,56----------->0,56
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)
0,04----------->0,08
\(\Rightarrow V_{CO_2}=\left(0,08+0,56\right).22,4=14,336\left(l\right)\)
check lại giúp mình câu b) nhé
C2H2 + Br2 ra 1,2-đibrometen(tỉ lệ 1:1) hoặc 1,1,2,2-tetrabrometan(tỉ lệ 1:2) chứ không ra đibrometan nhé :v