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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: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,0175\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,0175.22,4}{0,86}.100\%\approx45,58\%\)
\(\Rightarrow\%V_{CH_4}\approx54,42\%\)
a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(1\right)\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{48}{160}=0,3\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,2.22,4}{5,6}.100\%=80\%\\\%V_{C_2H_2}=20\%\end{matrix}\right.\)
b, \(V_{ddBr_2}=\dfrac{0,3}{0,25}=1,2\left(M\right)\)
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, \(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)
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.
\(V_{CH_4}=11,2.70\%=7,84\left(l\right)\)
a.b.\(m_{tăng}=m_{C_2H_4}=2,8g\)
\(n_{C_2H_4}=\dfrac{2,8}{28}=0,1mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(n_{hh}=\dfrac{6,72}{22,4}=0,3mol\)
\(\rightarrow m_{CH_4}=\left(0,3-0,1\right).16=3,2g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{2,8}{2,8+3,2}.100=46,67\%\\\%m_{CH_4}=100\%-46,67\%=53,33\%\end{matrix}\right.\)
c.\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,2 0,2 ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
0,1 0,2 ( mol )
\(V_{CO_2}=\left(0,2+0,2\right).22,4=8,96l\)
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.
C2H2+2Br2-to>C2H2Br4
0,4-------0,8----------0,4 mol
n C2H2Br4=\(\dfrac{138,4}{346}=0,4mol\)
=>m Br2= 0,8.160=128g
=>m ddBr2=1280g
=>%mC2H2=\(\dfrac{0,4.22,4}{16,8}.100=53,3\%\)
=>%m CH4=100-53,3=46,7%