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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
\(a,Gọi\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{C_2H_2}=c\left(mol\right)\end{matrix}\right.\\ n_{hhkhí}=0,4\left(mol\right)\\ n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:a\rightarrow2a\rightarrow a\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:b\rightarrow3b\rightarrow2b\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:c\rightarrow2,5c\rightarrow2c\\ Hệ.pt\left\{{}\begin{matrix}a+b+c=0,4\\b+2c=0,4\\a+2b+2c=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\)
\(\%V_{CH_4}=\%V_{C_2H_2}=\dfrac{0,1}{0,4}=25\%\\ \%V_{C_2H_4}=\dfrac{0,2}{0,4}=50\%\)
\(m_{CH_4}=0,1.16=1,6\left(g\right)\\ m_{C_2H_4}=28.0,2=5,6\left(g\right)\\ m_{C_2H_2}=0,1.26=2,6\left(g\right)\\ \%m_{CH_4}=\dfrac{1,6}{1,6+5,6+2,6}=16,32\%\\ \%m_{C_2H_4}=\dfrac{5,6}{1,6+5,6+2,6}=57,14\%\\ \%m_{C_2H_2}=100\%-16,32\%-57,14\%=26,54\%\)
\(b,PTHH:C_2H_5OH\rightarrow C_2H_4+H_2O\\ Mol:0,2\leftarrow0,2\\ m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)
Dài quá!!!
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: \(n_{hh}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,075}{0,225}\cdot100\%\approx33,33\%\\\%V_{C_2H_4}=\dfrac{0,1}{0,225}\cdot100\%\approx44,44\%\\\%V_{C_2H_2}=22,23\%\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(m_{tăng}=m_{Br_2}=m_{C_2H_2}=2,6g\)
\(\Rightarrow n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
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, nBr2 = 8/160 = 0,05 (mol)
PTHH: C2H4 + Br2 -> C2H4Br2
Mol: 0,05 <--- 0,05 <--- 0,05
Vhh khí = 2,8/22,4 = 0,125 (mol)
%VC2H4 = 0,05/0,125 = 40%
%CH4 = 100% - 40% = 60%
b, nCH4 = 0,125 - 0,05 = 0,075 (mol)
PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O
Mol: 0,05 ---> 0,15
CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,075 ---> 0,15
Vkk = (0,15 + 0,15) . 5 . 22,4 = 33,6 (l)
a) mtăng = mC2H4
=> \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)
=> \(\%V_{C_2H_4}=\dfrac{0,2.22,4}{13,44}.100\%=33,33\%\)
\(\%V_{CH_4}=100\%-33,33\%=66,67\%\)
b) \(n_{CH_4}=\dfrac{13,44.66,67\%}{22,4}=0,4\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4--------------->0,4
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,2----------------->0,4
Ca(OH)2 + CO2 --> CaCO3 + H2O
0,8----->0,8
=> mCaCO3 = 0,8.100 = 80 (g)
a.\(m_{tăng}=m_{C_2H_4}=5,6g\)
\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)
\(n_{C_2H_4}=\dfrac{5,6}{28}=0,2mol\)
\(\%V_{C_2H_4}=\dfrac{0,2}{0,6}.100=33,33\%\)
\(\%V_{CH_4}=100\%-33,33\%=66,67\%\)
b.\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
0,2 0,4 ( mol )
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,4 0,4 ( mol )
\(n_{CO_2}=0,4+0,4=0,8mol\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
0,8 0,8 ( mol )
\(m_{CaCO_3}=0,8.100=80g\)
Theo gt ta có: $n_{hh}=0,08(mol);n_{Br_2}=0,08(mol)$
$C_2H_2+2Br_2\rightarrow C_2H_2Br_4$
Suy ra $n_{C_2H_2}=0,04(mol)=n_{CH_4}$
a, $\Rightarrow \%V_{C_2H_2}=\%V_{C_2H_4}=50\%$
b, $CH_4+2O_2\rightarrow CO_2+2H_2O$
$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$
Ta có: $n_{O_2}=0,04.2+0,04.5=0,28(mol)\Rightarrow m_{O_2}=8,96(g)$
\(a)C_2H_2 +2Br_2 \to C_2H_2Br_2\\ n_{C_2H_2} = \dfrac{1}{2}n_{Br_2} = \dfrac{0,4.0,2}{2} = 0,04(mol)\\ \Rightarrow V_{C_2H_2} = 0,04.22,4 = 0,896(lít)\\ \%V_{C_2H_2} =\dfrac{0,896}{1,792}.100\% = 50\%\\ \Rightarrow \%V_{CH_4} = 100\% -50\% = 50\%\\ b)\\V_{CH_4} = V_{C_2H_2} = 0,896(lít)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_2 + \dfrac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + H_2O\\ \)
\(V_{O_2} = 2V_{CH_4} + \dfrac{5}{2}V_{C_2H_2} = 4,032(lít)\\ \Rightarrow m_{O_2} = \dfrac{4,032}{22,4}.32 = 5,76(gam)\)