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a, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, - Khí thoát ra là CH4.
⇒ VCH4 = 4,48 (l)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{4,48}{11,2}.100\%=40\%\\\%V_{C_2H_4}=100-40=60\%\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right);n_{CH_4}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ a,CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ Đặt:n_{CH_4}=a\left(mol\right);n_{C_2H_4}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+2b=0,4\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow\%V_{CH_4}=\%n_{CH_4}=\dfrac{a}{0,3}.100\%=\dfrac{0,2}{0,3}.100\approx66,667\%\\ \Rightarrow\%V_{C_2H_4}\approx33,333\%\\ c,C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\\ \Rightarrow C_{MddBr_2}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
Chỗ kia chắc 200ml dung dịch Br2 chứ 200gam thì cần cho thêm KLR á em
a) C2H4 + Br2 --> C2H4Br2
b) nBr2 = 2.0,15 = 0,3 (mol)
PTHH: C2H4 + Br2 --> C2H4Br2
0,3<-- 0,3----->0,3
=> \(m_{C_2H_4Br_2}=0,3.188=56,4\left(g\right)\)
c) \(\%V_{C_2H_4}=\dfrac{0,3.22,4}{22,4}.100\%=30\%\)
\(\%V_{CH_4}=100\%-30\%=70\%\)
m(tăng) = mC2H4 (tham gia p/ư) = 2,8 (g)
nC2H4 = 2,8/28 = 0,1 (mol)
VC2H4 = 0,1 . 22,4 = 2,24 (l)
VCH4 = 4,48 - 2,24 = 2,24 (l)
1. \(n_{Br_2}=0,4.0,5=0,2\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)
2. \(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,05\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,05.22,4}{1,4}.100\%=80\%\)
\(\Rightarrow\%V_{CH_4}=100-80=20\%\)
a) nC2H4Br2=47/188=0,25(mol)
n(CH4,C2H4)=11,2/22,4=0,5(mol)
PTHH: C2H4 + Br2 -> C2H4Br2
0,25<----------0,25<---------0,25(mol)
mBr2(p.ứ)=0,25 x 160= 40(g)
b) V(C2H4,đktc)=0,25 x 22,4= 5,6(l)
=> %V(C2H4)=(5,6/11,2).100=50%
=>%V(CH4)=100% - 50%= 50%
a) C2H4 + Br2 --> C2H4Br2
b) nBr2 = 0,2.0,2 = 0,04 (mol)
PTHH: C2H4 + Br2 --> C2H4Br2
0,04<--0,04
=> \(m_{C_2H_4}=0,04.28=1,12\left(g\right)\)
\(m_{CH_4}=n_{CH_4}.M_{CH_4}=\left(\dfrac{1,12}{22,4}-0,04\right).16=0,16\left(g\right)\)
c) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,04.22,4}{1,12}.100\%=80\%\\\%V_{CH_4}=100\%-80\%=20\%\end{matrix}\right.\)
\(n_{hh}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{C_2H_4Br_2}=\dfrac{16}{188}=\dfrac{4}{47}mol\)
\(\Rightarrow n_{etilen}=\dfrac{4}{47}mol\)
\(\Rightarrow n_{metan}=0,3-\dfrac{4}{47}=\dfrac{101}{470}mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(\%V_{etilen}=\dfrac{\dfrac{4}{47}}{0,3}\cdot100\%=28,37\%\)
\(\%V_{metan}=100\%-28,37\%=71,63\%\)
a) PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Hiện tượng: Dung dịch Brom bị nhạt màu
b) Ta có: \(n_{Br_2}=\dfrac{24}{160}=0,15\left(mol\right)=n_{C_2H_4Br_2}\)
\(\Rightarrow m_{C_2H_4Br_2}=0,15\cdot188=28,2\left(g\right)\)
c) Theo PTHH: \(n_{C_2H_4}=0,15\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,15\cdot22,4}{4,48}\cdot100\%=75\%\)
\(\Rightarrow\%V_{CH_4}=25\%\)