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n Br2=\(\dfrac{32}{160}\)=0,2 mol
C2H2+2Br2->C2H2Br4
0,1------0,2 mol
=>%VC2H2=\(\dfrac{0,1.22,4}{5,6}\).100=40%
=>%VCH4=100-40=60%
=>n CH4=\(\dfrac{5,6-0,1.22,4}{22,4}\)=0,15 mol
CH4+2O2-to>CO2+2H2O
0,15----0,3
C2H2+\(\dfrac{5}{2}\)O2-to>2CO2+H2O
0,1-----0,25 mol
=>VO2=(0,3+0,25).22,4=12,32l
a) \(C_2H_4 + Br_2 \to C_2H_4Br_2\)
b)
\(n_{C_2H_4} = n_{Br_2} = \dfrac{5,6}{160} = 0,035(mol)\\ \%V_{C_2H_4} = \dfrac{0,035.22,4}{0,86}.100\% = 91,16\%\\ \%V_{CH_4} = 100\% - 91,16\% = 8,84\%\)
a, \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{72}{160}=0,45\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,15.28}{0,15.28+0,15.26}.100\%\approx51,85\%\\\%m_{C_2H_2}\approx48,15\%\end{matrix}\right.\)
\(\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,15.22,4}{6,72}.100\%=50\%\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{C_2H_4Br_2}=n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2Br_4}=n_{C_2H_2}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{C_2H_4Br_2}=0,15.188=28,2\left(g\right)\\m_{C_2H_2Br_4}=0,15.346=51,9\left(g\right)\end{matrix}\right.\)
Ta có: \(n_{C_2H_2Br_2}=\dfrac{173}{186}\left(mol\right)\)
PT: \(C_2H_2+Br_2\rightarrow C_2H_2Br_2\)
Theo PT: \(n_{C_2H_2}=n_{C_2H_2Br_2}=\dfrac{173}{186}\left(mol\right)\)
\(\Rightarrow V_{C_2H_2}=\dfrac{173}{186}.22,4=20,83\left(l\right)\) > Vhh → vô lý
Bạn xem lại đề nhé.
a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{C_2H_4}=n_{Br_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{3,36}.100\%\approx66,67\%\\\%V_{CH_4}\approx33,33\%\end{matrix}\right.\)
a)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,05<-0,05
=> \(n_{CH_4}=\dfrac{3,36}{22,4}-0,05=0,1\left(mol\right)\)
\(\%m_{CH_4}=\dfrac{0,1.16}{0,1.16+0,05.28}.100\%=53,33\%\)
\(\%m_{C_2H_4}=\dfrac{0,05.28}{0,1.16+0,05.28}.100\%=46,67\%\)
b)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,1-->0,2
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,05--->0,15
=> \(V_{O_2}=\left(0,2+0,15\right).22,4=7,84\left(l\right)\)
Gọi số mol C2H4, C2H2 là a, b (mol)
=> a + b = \(\dfrac{0,672}{22,4}=0,03\left(mol\right)\) (1)
PTHH: C2H4 + Br2 --> C2H4Br2
a-------------->a
C2H2 + 2Br2 --> C2H2Br4
b----------------->b
=> 188a + 346b = 8,8 (2)
(1)(2) => a = 0,01 (mol); b = 0,02 (mol)
=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,01}{0,03}.100\%=33,33\%\\\%V_{C_2H_2}=\dfrac{0,02}{0,03}.100\%=66,67\%\end{matrix}\right.\)