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a, Ta có \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,05\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,05.22,4=1,12\left(l\right)\\V_{C_2H_4}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)
b, \(m_{CH_4}=0,05.16=0,8\left(g\right)\)
c, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)
\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
a) Ta có: \(n_{C_2H_4}=\dfrac{9,1}{28}=0,325\left(mol\right)=n_{Br_2}\) \(\Rightarrow V_{Br_2}=\dfrac{0,325}{2}=0,1625\left(l\right)=162,5\left(ml\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{C_2H_4}=0,325\left(mol\right)\\n_{CH_4}=\dfrac{13,44}{22,4}-0,325=0,275\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{hh}=9,1+0,275\cdot16=13,5\left(g\right)\)
c) PTHH: \(CH_4+2O_2 \underrightarrow{t^o} CO_2+2H_2O\)
\(C_2H_4+3O_2 \underrightarrow{t^o} 2CO_2+ 2H_2O\)
Theo các PTHH: \(\Sigma n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=1,525\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,525\cdot22,4=34,16\left(l\right)\)
\(n_{hhk}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(n_{C_2H_4Br_2}=\dfrac{47}{188}=0,25\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,25 0,25 0,25 ( mol )
\(\%V_{C_2H_4}=\dfrac{0,25}{0,4}.100=62,5\%\)
\(\%V_{CH_4}=100-62,5=37,5\%\)
\(V_{Br_2}=\dfrac{0,25}{1}=0,25\left(l\right)\)
\(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\)\(\Rightarrow a + b = \dfrac{3,36}{22,4} = 0,15(1) \)
\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ n_{CO_2} = a + 2b = \dfrac{4,48}{22,4} = 0,2(2)\)
Từ (1)(2) suy ra: a = 0,1 ; b = 0,05
Suy ra:
\(\%V_{CH_4} = \dfrac{0,1}{0,15}.100\% = 66,67\%\\ \%V_{C_2H_4} = 100\% - 66,67\% = 33,33\%\)
b)
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{Br_2} = n_{C_2H_4} = 0,05(mol)\\ \Rightarrow m_{Br_2} = 0,05.160 = 8\ gam\)
Gọi số mol C2H4, C2H2 là a, b (mol)
=> a + b = \(\dfrac{13,44}{22,4}=0,6\) (1)
nBr2 = 0,8.1 = 0,8 (mol)
PTHH: C2H4 + Br2 --> C2H4Br2
a----->a
C2H2 + 2Br2 --> C2H2Br4
b----->2b
=> a + 2b = 0,8 (2)
(1)(2) => a = 0,4 (mol); b = 0,2 (mol)
=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,4}{0,6}.100\%=66,67\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,6}.100\%=33,33\%\end{matrix}\right.\)
nhh = 6.72/22.4 = 0.3 (mol)
nBr2 = 64/160 = 0.4 (mol)
nC2H4 = a (mol) . nC2H2 = b (mol)
C2H2 + 2Br2 => C2H2Br4
C2H4 + Br2 => C2H4Br2
=> a + b = 0.3
a + 2b = 0.4
=> a =0.2 , b = 0.1
%VC2H4 = 0.2/0.3 * 100% = 66.67%
%VC2H2 = 33.33%
a, vì CH4 là hidrocacbon no => không xảy ra phản ứng với Brom
pt: C2H4 + Br2 -> C2H4Br2
1 1 1
nBr2 = m/M = 6,4/160 = 0,04 mol => nC2H4 = 0,04 mol
=> VC2H4 = n x 22,4 = 0,04 x 22,4 = 0,896 lit
=> VCH4 = Vhh - VC2H4 = 6,72 - 0,896 = 5,824 lit
b, C%VC2H4 = VC2H4/Vhh = 0,896/6,72 X 100 = 13,33%
=> C%VCH4 = Vhh - VC2H4 = 100% - 13,33% = 86,67%
\(n_{hh}=6,72:22,4=0,3mol\\ C_2H_2+2Br_2->C_2H_2Br_4\\ C_2H_4+Br_2->C_2H_2Br_2\\ n_{Br_2}=0,4mol\\ n_{C_2H_2}=a;n_{C_2H_4}=b\\ a+b=0,3\\ 2a+b=0,4\\ a=0,2;b=0,1\\ \%V_{C_2H_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\ \%V_{C_2H_4}=33,33\%\)
1/2 hỗn hợp có 0,1 mol C2H2 và 0,05mol C2H4
\(BT.C:n_{CO_2}=2n_{C_2H_2}+2n_{C_2H_4}=0,3mol\\ n_{CaCO_3}=n_{CO_2}=0,3\\ m_{KT}=0,3.100=30g\)