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a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
b, - Khí thoát ra là CH4 ⇒ VCH4 = 6,72 (l)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{6,72}{13,44}.100\%=50\%\\\%V_{C_2H_2}=50\%\end{matrix}\right.\)
a. PTHH: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
a. Vì CH4 không phản ứng với dd Br2 nên
\(V_{CH_4}=6,72\left(l\right)\)
\(\%V_{CH_4}=\dfrac{6,72}{13,44}x100\%=50\%\)
\(\%V_{C_2H_2}=100\%-50\%=50\%\)
\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{16}{160}=0,1mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,1 0,1 ( mol )
\(\%V_{C_2H_4}=\dfrac{0,1.22,4}{16,8}.100=13,33\%\)
\(\%V_{CH_4}=100\%-13,33\%=86,67\%\)
a, PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
b, \(n_{Br_2}=\dfrac{5,6}{160}=0,035\left(mol\right)\)
Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,0175\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,0175.22,4}{0,86}.100\%\approx45,58\%\)
\(\Rightarrow\%V_{CH_4}\approx54,42\%\)
\(n_{Br_2}=\dfrac{20}{160}=0,125\left(mol\right)\)
PTHH: C2H4 + Br2 ---> C2H4Br2
0,125 0,125
\(\%V_{C_2H_4}=\dfrac{0,125.22,4}{5,6}=50\%\\ \%V_{CH_4}=100\%-50\%=50\%\)
a, \(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)
PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
_____0,15____0,3 (mol)
\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,15.22,4}{11,2}.100\%=30\%\)
\(\Rightarrow\%V_{CH_4}=100-30=70\%\)
b, - Khí thoát ra ngoài là CH4.
\(V_{CH_4}=11,2.70\%=7,84\left(l\right)\)
\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{Br_2}=\dfrac{2,4}{160}=0,015mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,0075 0,015 ( mol )
\(V_{C_2H_2}=0,0075.22,4=0,168l\)
\(V_{CH_4}=3,36-0,168=3,192l\)
\(\%V_{C_2H_2}=\dfrac{0,168}{3,36}.100=5\%\)
\(\%V_{CH_4}=100\%-5\%=95\%\)
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)\(\%V_{metan}=\dfrac{0,2}{0,25}\cdot100\%=80\%\)
\(\%V_{etilen}=100\%-80\%=20\%\)
\(n_{hhk}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
a, PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
b, Ta có: \(n_{Br_2}=\dfrac{4,8}{160}=0,03\left(mol\right)\)
Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,015\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,015}{0,125}.100\%=12\%\\\%V_{CH_4}=88\%\end{matrix}\right.\)
c, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Ta có: \(n_{CH_4}=0,125-0,015=0,11\left(mol\right)\)
Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,2575\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,2575.22,4=5,768\left(l\right)\)
Mà: VO2 = 20% Vkk
\(\Rightarrow V_{kk}=5,768.5=28,84\left(l\right)\)
Bạn tham khảo nhé!