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\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\) ( mol ) \(\Rightarrow m_{hh}=16x+28y=6\left(g\right)\) (1)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x x ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
y 2y ( mol )
\(n_{CO_2}=x+2y=0,4\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%m_{CH_4}=\dfrac{0,2.16}{6}.100=53,33\%\)
\(\%m_{C_2H_4}=100-53,33=46,67\%\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,1 0,1 ( mol )
\(m_{Br_2}=0,1.160=16\left(g\right)\)
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: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)
\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)
a)
\(C_2H_4 + Br_2 \to C_2H_4Br_2 n_{C_2H_4} = n_{Br_2} = \dfrac{8}{160} = 0,05(mol)\\ \Rightarrow \%V_{C_2H_4} = \dfrac{0,05.22,4}{2,24} .100\%= 50\%\\ \%V_{C_2H_4} = 100\%-50\% = 50\%\)
b)
\(C_2H_4 + H_2 \xrightarrow{t^o,Ni} C_2H_6\\ n_{H_2\ pư} = n_{C_2H_6} = \dfrac{0,896}{22,4} = 0,04(mol)\\ \Rightarrow m_{H_2\ pư} = 0,04.2 = 0,08(gam)\)
\(CO+CuO\underrightarrow{t^0}Cu+CO_2\)
Hỗn hợp A gồm Cu và CuO dư sau phản ứng
\(Cu+2H_2SO_{4\left(đ,t^0\right)}\rightarrow CuSO_4+SO_2+2H_2O\)
\(CuO+H_2SO_{4\left(đ,t^0\right)}\rightarrow CuSO_4+H_2O\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
\(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\)\(\Rightarrow a + b = \dfrac{3,36}{22,4} = 0,15(1) \)
\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ n_{CO_2} = a + 2b = \dfrac{4,48}{22,4} = 0,2(2)\)
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)
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{Br_2} = n_{C_2H_4} = 0,05(mol)\\ \Rightarrow m_{Br_2} = 0,05.160 = 8\ gam\)
a) \(V_{CH_4}=0,6\left(l\right)\)
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,6}{1,2}.100\%=50\%\\\%V_{C_2H_4}=100\%-50\%=50\%\end{matrix}\right.\)
b) \(n_{C_2H_4}=\dfrac{1,2-0,6}{24}=0,025\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,025-->0,025
=> \(m_{Br_2}=0,025.160=4\left(g\right)\)
c)
\(n_{CH_4}=\dfrac{0,6}{24}=0,025\left(mol\right)\)
=> nH = 0,025.4 = 0,1 (mol)
\(n_{Cl_2}=\dfrac{0,72}{24}=0,03\left(mol\right)\)
=> nCl(thế H) = 0,03 (mol)
Do nH > nCl(thế H)
=> H không bị thế hoàn toàn bởi Cl
=> nHCl = 0,03 (mol)
=> mHCl = 0,03.36,5 = 1,095 (g)
\(n_{Br_2}=\dfrac{16}{160}=0,1mol\Rightarrow n_{CH_2}=0,1mol\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(\Rightarrow n_{CH_4}=0,25-0,1=0,15mol\)
\(\%V_{CH_2}=\dfrac{0,1}{0,25}\cdot100\%=40\%\)
\(\%V_{CH_4}=100\%-40\%=60\%\)
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(CH_2+\dfrac{3}{2}O_2\underrightarrow{t^o}CO_2+H_2O\)
\(\Rightarrow\Sigma n_{CO_2}=0,15+0,1=0,25mol\)
\(BTC:n_{CO_2}=n_{CaCO_3}=0,25mol\)
\(\Rightarrow m_{\downarrow}=0,25\cdot100=25g\)
A. Dẫn hh khí qua dd Brom dư