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a, Ta có: \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)
\(16n_{CH_4}+28n_{C_2H_4}=3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,1\left(mol\right)\\n_{C_2H_4}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%\approx53,33\%\\\%m_{C_2H_4}\approx46,67\%\end{matrix}\right.\)
- Ở cùng điều kiện nhiệt độ và áp suất, % số mol cũng là %V.
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)
b, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Có: m tăng = mC2H4 = 0,05.28 = 1,4 (g)
a) \(n_{hh}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=0,15\\16a+28b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%=66,67\%\\\%V_{C_2H_4}=100\%-66,67\%=33,33\%\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%=53,33\%\\\%m_{C_2H_4}=100\%-53,33\%=46,67\%\end{matrix}\right.\)
b) \(m=m_{C_2H_4}=0,05.28=1,4\left(g\right)\)
ta có :
nBr2=\(\dfrac{16}{160}=0,1mol\)
C2H4+Br2->C2H4Br2
0,1------0,1
=>VC2H4=0,1.22,4=2,24l
=>VCH4=3,36l->n CH4=0,15 mol
->%VC2H4=\(\dfrac{2,24}{5,6}.100\)=40%
=>%VCH4=60%
c)
CH4+2O2-to>CO2+2H2O
0,15---------------0,15
C2H4+3O2--to>2CO2+2H2O
0,1--------------------0,2
=>m CaCO3=0,35.100=35g
\(n_{Br_2}=\dfrac{8}{160}=0.05\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(0.05......0.05\)
\(V_{C_2H_4}=0.05\cdot22.4=1.12\left(l\right)\)
\(V_{CH_4}=20-1.12=18.88\left(l\right)\left(mol\right)\)
\(\%V_{C_2H_4}=\dfrac{1.12}{20}\cdot100\%=5.6\%\)
\(\%V_{CH_4}=100-5.6=94.4\%\)
a)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,05<--0,05
=> \(V_{C_2H_4}=0,05.22,4=1,12\left(l\right)\)
=> \(V_{CH_4}=4,48-1,12=3,36\left(l\right)\)
b) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{1,12}{4,48}.100\%=25\%\\\%V_{CH_4}=\dfrac{3,36}{4,48}.100\%=75\%\end{matrix}\right.\)
a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{C_2H_4}=n_{Br_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{3,36}.100\%\approx66,67\%\\\%V_{CH_4}\approx33,33\%\end{matrix}\right.\)
`C_2H_4+Br_2->C_2H_4Br_2`
`0,09---0,09mol`
`n(Br_2)=(14,4)/160=0,09mol`
`->%(C_2H_4)=((0,09.22,4)/(3,36)).100%=60%`
`->%(CH_4)=100-60=40%`
#yBTrCute