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a) Gọi số mol CH4, C2H4 là a, b (mol)
=> \(\left\{{}\begin{matrix}16a+28b=11,6\\a+b=\dfrac{11,2}{22,4}=0,5\end{matrix}\right.\)
=> a = 0,2 (mol); b = 0,3 (mol)
\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2}{0,5}.100\%=40\%\\\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,2.16}{11,6}.100\%=27,586\%\\\%m_{C_2H_4}=\dfrac{0,3.28}{11,6}.100\%=72,414\%\end{matrix}\right.\)
b)
\(n_{C_2H_4}=\dfrac{5,6.60\%}{22,4}=0,15\left(mol\right)\)
mtăng = mC2H4 = 0,15.28 = 4,2 (g)
\(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\)\(\Rightarrow a + b = \dfrac{3,36}{22,4} = 0,15(1) \)
\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ n_{CO_2} = a + 2b = \dfrac{4,48}{22,4} = 0,2(2)\)
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, 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: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)
\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)
$C_2H_4 + Br_2 \to C_2H_4Br_2$
Ta có :
$m_{C_2H_4} = m_{dd\ tăng} = 3(gam)$
$\Rightarrow n_{C_2H_4} = \dfrac{3}{28}(mol)$
$\Rightarrow n_{C_2H_6} = 0,25 - \dfrac{3}{28} = \dfrac{1}{7}(mol)$
$\%V_{C_2H_4} = \dfrac{ \dfrac{3}{28} }{0,25}.100\% = 42,9\%4
$\%V_{C_2H_6} = 100\% - 42,9\% = 57,1\%$
$\%m_{C_2H_4} = \dfrac{3}{3 + \dfrac{1}{7}.30}.100\% = 41,2\%$
$\%m_{C_2H_6} = 100\% - 41,2\% = 58,8\%$
$C_2H_4 + Br_2 \to C_2H_4Br_2$
Ta có :
$m_{C_2H_4} = m_{dd\ tăng} = 3(gam)$
$\Rightarrow n_{C_2H_4} = \dfrac{3}{28}(mol)$
$\Rightarrow n_{C_2H_6} = 0,25 - \dfrac{3}{28} = \dfrac{1}{7}(mol)$
$\%V_{C_2H_4} = \dfrac{ \dfrac{3}{28} }{0,25}.100\% = 42,9\%$
$\%V_{C_2H_6} = 100\% - 42,9\% = 57,1\%$
$\%m_{C_2H_4} = \dfrac{3}{3 + \dfrac{1}{7}.30}.100\% = 41,2\%$
$\%m_{C_2H_6} = 100\% - 41,2\% = 58,8\%$
\(a,n_{hh\left(CH_4,C_2H_4,C_2H_2\right)}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{hh\left(C_2H_4,C_2H_2\right)}=0,4-0,1=0,3\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a+b=0,3\\28a+26b=8,1\end{matrix}\right.\Leftrightarrow a=b=0,15\left(mol\right)\)
PTHH:
\(CH\equiv CH+2Br-Br\rightarrow CHBr_2-CHBr_2\)
\(CH_2=CH_2+Br-Br\rightarrow CH_2Br-CH_2Br\)
\(b,\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,4}.100\%=25\%\\\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,15}{0,4}.100\%=37,5\%\end{matrix}\right.\)
c, PTHH:
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\\ C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\\ \rightarrow n_{BaCO_3}=n_{CO_2}=0,1+0,15.0,15.2=0,7\left(mol\right)\\ m_{BaCO_3}=0,7.197=137,9\left(g\right)\)
a) PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Ta có: \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)=n_{C_2H_4Br_2}\) \(\Rightarrow m_{C_2H_4Br_2}=0,2\cdot188=37,6\left(g\right)\)
b) Ta có: \(n_{CH_4}=\dfrac{11,2}{22,4}-0,2=0,3\left(mol\right)\)
Bảo toàn nguyên tố: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,7\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,7\cdot22,4=15,68\left(l\right)\)
a) mtăng = mC2H4
=> \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)
=> \(\%V_{C_2H_4}=\dfrac{0,2.22,4}{13,44}.100\%=33,33\%\)
\(\%V_{CH_4}=100\%-33,33\%=66,67\%\)
b) \(n_{CH_4}=\dfrac{13,44.66,67\%}{22,4}=0,4\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4--------------->0,4
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,2----------------->0,4
Ca(OH)2 + CO2 --> CaCO3 + H2O
0,8----->0,8
=> mCaCO3 = 0,8.100 = 80 (g)
a.\(m_{tăng}=m_{C_2H_4}=5,6g\)
\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)
\(n_{C_2H_4}=\dfrac{5,6}{28}=0,2mol\)
\(\%V_{C_2H_4}=\dfrac{0,2}{0,6}.100=33,33\%\)
\(\%V_{CH_4}=100\%-33,33\%=66,67\%\)
b.\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
0,2 0,4 ( mol )
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,4 0,4 ( mol )
\(n_{CO_2}=0,4+0,4=0,8mol\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
0,8 0,8 ( mol )
\(m_{CaCO_3}=0,8.100=80g\)
\(\left\{{}\begin{matrix}CH_4:x\left(mol\right)\\C_2H_2:y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}16x+26y=11,6\\x+y=\dfrac{11,2}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,14mol\\y=0,36mol\end{matrix}\right.\)
a)\(\%V_{CH_4}=\dfrac{0,14}{0,5}\cdot100\%=28\%\)
\(\%V_{C_2H_2}=100\%-28\%=72\%\)
\(\%m_{CH_4}=\dfrac{0,14\cdot16}{11,6}\cdot100\%=19,31\%\)
\(\%m_{C_2H_2}=100\%-19,31\%=80,69\%\)
b)\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
Dẫn dung dịch qua bình đựng brom chỉ có \(C_2H_2\) tác dụng.
\(\Rightarrow m_{tăng}=m_{C_2H_2}=0,25\cdot26=6,5g\)