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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: \(n_{hh}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,075}{0,225}\cdot100\%\approx33,33\%\\\%V_{C_2H_4}=\dfrac{0,1}{0,225}\cdot100\%\approx44,44\%\\\%V_{C_2H_2}=22,23\%\end{matrix}\right.\)
nhh = 5.6/22.4 = 0.25 (mol)
nBr2 = 0.5*0.2 = 0.1 (mol)
C2H4 + Br2 => C2H4Br2
0.1_____0.1
nCH4 = 0.25 - 0.1 = 0.15 (mol)
VC2H4 = 0.1*22.4 = 2.24 (l)
VCH4 = 0.15*22.4 = 3.36 (l)
mCH4 = 0.15*16 = 2.4 (g)
mC2H4 = 0.1*28 = 2.8 (g)
%mCH4 = 2.4/(2.4 + 2.8) * 100% = 46.15%
%mC2H4 = 100 - 46.15 = 53.85%
Chúc bạn học tốt !!!
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(m_{tăng}=m_{Br_2}=m_{C_2H_2}=2,6g\)
\(\Rightarrow n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
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)\(\Sigma n_{O_2}=0,4+0,25=0,65mol\)
\(\Rightarrow V_{O_2}=0,65\cdot22,4=14,56l\)
\(\Rightarrow V_{kk}=14,56\cdot5=72,8l\)
\(m_{tăng}=m_{C_2H_2}=1,3\left(g\right)\\ \Rightarrow n_{C_2H_2}=\dfrac{1,3}{26}=0,05\left(mol\right)\\ \Rightarrow\%V_{\dfrac{C_2H_2}{A}}=\dfrac{0,05.22,4}{4,48}.100=25\%\\ \Rightarrow\%V_{\dfrac{CH_4}{A}}=100\%-25\%=75\%\)
a)
\(m_{C_2H_2} = m_{tăng} = 5,2\ gam\\ \Rightarrow n_{C_2H_2} = \dfrac{5,2}{26} = 0,2(mol)\)
Vậy :
\(\%V_{C_2H_2} = \dfrac{0,2.22,4}{8,96}.100\% = 50\%\\ \%V_{CH_4} = 100\%-50\% = 50\%\)
b)
\(n_{CH_4} = n_{C_2H_2} = 0,2(mol)\)
CH4 + O2 \(\xrightarrow{t^o}\) CO2 + H2O
0,2.........................0,2...................................(mol)
C2H2 + \(\dfrac{5}{2}\)O2 \(\xrightarrow{t^o}\) 2CO2 + H2O
0,2................................0,4.................................(mol)
CO2 + Ca(OH)2 → CaCO3 + H2O
(0,2+0,4)............................(0,2+0,4)........................................(mol)
\(\Rightarrow m_{CaCO_3} =(0,2 + 0,4).100 = 60(gam)\)
a) \(n_{CH_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
=> \(\%V_{CH_4}=\dfrac{3,36}{11,2}.100\%=30\%\)
=> \(\%V_{C_2H_4}=100\%-30\%=70\%\)
b) \(n_{C_2H_4}=\dfrac{11,2.70\%}{22,4}=0,35\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,15-->0,3
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,35-->1,05
=> nO2 = 0,3 + 1,05 = 1,35 (mol)
=> VO2 = 1,35.22,4 = 30,24 (l)
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) \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{\dfrac{2016}{65}}{\dfrac{2016}{65}+8,4}.100\%=78,69\%\\\%V_{CH_4}=\dfrac{8,4}{\dfrac{2016}{65}+8,4}.100\%=21,31\%\end{matrix}\right.\)
C2H2 + 2Br2 => C2H2Br4
CH4 + 2O2 -to-> CO2 + 2H2O
m bình tăng = mC2H2 = 5.2 (g)
nC2H2 = 5.2/26 = 0.2 (mol)
nCO2 = 13.2/44 = 0.3 (mol) => nCH4 = 0.3 (mol)
Vì : tỉ lệ thể tích tương ứng với tỉ lệ số mol
%C2H2 = 0.2/(0.2+0.3) * 100% = 40%
%CH4 = 100 - 40 = 60%
nBr2 = 2nC2H2 = 0.2*2 = 0.4 (mol)
CM Br2 = 0.4/0.4 = 1M