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nC2H4Br2 = \(\dfrac{4,7}{188}\)=0,025(mol)
C2H4 + Br2 -> C2H4Br2
0,025 <-----------0,025
=>VC2H4 = 0,025 . 22,4=0,56(l)
=> VCH4 = 2,8 - 0,56 =2,24 (l)
%VCH4 =\(\dfrac{2,24.100}{2,8}\)=80%
%VC2H4 = 100 % -80% = 20%
Bài 9 :
Metan không tác dụng với dung dịch Brom nên :
\(n_{C2H4Br2}=\dfrac{4,7}{188}=0,025\left(mol\right)\)
a) Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_2|\)
1 1 1
0,025 0,025
b) \(n_{C2H4}=\dfrac{0,025.1}{1}=0,025\left(mol\right)\)
\(V_{C2H4\left(dktc\right)}=0,025.22,4=0,56\left(l\right)\)
\(\%V_{C2H4}=\dfrac{0,56.100}{2,8}=20\%\)
\(\%V_{CH4}=100\%-20\%=80\%\)
Chúc bạn học tốt
\(n_{C_2H_2Br_4}=\dfrac{17.3}{346}=0.05\left(mol\right)\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{C_2H_2}=0.05\left(mol\right)\)
\(m_{C_2H_2}=0.05\cdot26=1.3\left(g\right)\)
\(m_{CH_4}=4.33-1.3=3.03\left(g\right)\)
\(\%m_{C_2H_2}=\dfrac{1.3}{4.33}\cdot100\%=30.02\%\)
\(\%m_{CH_4}=69.98\%\)
\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{16}{160}=0,1mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,1 0,1 ( mol )
\(\%V_{C_2H_4}=\dfrac{0,1.22,4}{16,8}.100=13,33\%\)
\(\%V_{CH_4}=100\%-13,33\%=86,67\%\)
Cho hỗn hợp qua dung dịch brom dư
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Khí thoát ra là \(CH_4\)
\(CH_4+2O_2\rightarrow^{t^o}CO_2+2H_2O\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
Ta có:
\(n_{CaCO_3}=\frac{40}{100}=0,4mol=n_{CO_2}=n_{CH_4}\)
\(\rightarrow V_{CH_4}=0,4.22,4=8,96l\)
\(\rightarrow\%V_{CH_4}=\frac{8,96}{13,56}=66\%\rightarrow\%V_{C_2H_4}=34\%\)
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)
Bài 14 :
Vì metan không tác dụng với Brom nên :
\(n_{C2H4Br2}=\dfrac{4,7}{188}=0,025\left(mol\right)\)
a) Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_{2|}\)
1 1 1
0,025 0,025
b) \(n_{C2H4}=\dfrac{0,025.1}{1}=0,025\left(mol\right)\)
\(V_{C2H4\left(dktc\right)}=0,025.22,4=0,56\left(l\right)\)
\(V_{CH4\left(dktc\right)}=1,4-0,56=0,84\left(l\right)\)
0/0VCH4 = \(\dfrac{0,84.100}{1,4}=60\)0/0
0/0VC2H4 = \(\dfrac{0,56.100}{1,4}=40\)0/0
Chúc bạn học tốt
\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{Br_2}=\dfrac{2,4}{160}=0,015mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,0075 0,015 ( mol )
\(V_{C_2H_2}=0,0075.22,4=0,168l\)
\(V_{CH_4}=3,36-0,168=3,192l\)
\(\%V_{C_2H_2}=\dfrac{0,168}{3,36}.100=5\%\)
\(\%V_{CH_4}=100\%-5\%=95\%\)
\(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05mol\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,05 0,05 0,05 ( mol )
\(m_{Br_2}=0,05.160=8g\)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,25}.100=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)
\(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right);n_{hh}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,025<-0,125
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,025}{0,125}.100\%=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)