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\(n_{O_2}=\dfrac{89.6}{22.4}=4\left(mol\right)\)
\(n_{H_2O}=3a\left(mol\right)\)
\(n_{CO_2}=a\left(mol\right)\)
\(2H_2+O_2\underrightarrow{^{^{t^0}}}2H_2O\)
\(2CO+O_2\underrightarrow{^{^{t^0}}}2CO_2\)
\(n_{O_2}=1.5a+0.5a=4\left(mol\right)\)
\(\Leftrightarrow a=2\)
\(n_{H_2}=3\left(mol\right),n_{CO}=1\left(mol\right)\)
\(\%V_{H_2}=\dfrac{3}{4}\cdot100\%=75\%\)
\(\%V_{CO}=25\%\)
\(\%m_{H_2}=\dfrac{3\cdot2}{3\cdot2+1\cdot28}\cdot100\%=17.64\%\)
\(\%m_{CO}=100-17.64=82.36\%\)
a)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,1<-0,05<-------0,1
2CO + O2 --to--> 2CO2
0,2<--0,1-------->0,2
=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\V_{CO}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)
b) \(m_{CO_2}=0,2.44=8,8\left(g\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1mol\)
\(2CO+O_2\rightarrow2CO_2\)
a 0,5a a
\(2H_2+O_2\rightarrow2H_2O\)
0,1 0,05 \(\leftarrow\) 0,1
\(\Sigma n_{O_2}=0,5a+0,05=0,15\)
\(\Rightarrow a=n_{O_2\left(CO\right)}=0,2mol\)
\(V_{CO}=2\cdot0,2\cdot22,4=8,96l\)
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{CO_2}=0,2\cdot44=8,8g\)
a)
2CO + O2 --to--> 2CO2
2H2 + O2 --to--> 2H2O
b) \(n_{H_2O}=\dfrac{12,6}{18}=0,7\left(mol\right)\); \(n_{CO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2CO + O2 --to--> 2CO2
0,6<--0,3<------0,6
2H2 + O2 --to--> 2H2O
0,7<--0,35<------0,7
=> \(\left\{{}\begin{matrix}V_{CO}=0,6.22,4=13,44\left(l\right)\\V_{H_2}=0,7.22,4=15,68\left(l\right)\end{matrix}\right.\)
VO2 = (0,3 + 0,35).22,4 = 14,56 (l)
c) \(M_A=\dfrac{0,6.28+0,7.2}{0,6+0,7}=14\left(g/mol\right)\)
=> \(d_{A/O_2}=\dfrac{14}{32}=0,4375\)
a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\) (1)
\(4H_2+O_2\underrightarrow{t^o}2H_2O\) (2)
b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)
c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)
a/ \(2CO\left(0,2\right)+O_2\left(0,1\right)\rightarrow2CO_2\left(0,2\right)\)
\(2H_2\left(0,1\right)+O_2\left(0,05\right)\rightarrow2H_2O\left(0,1\right)\)
\(n_{H_2O}=\frac{1,8}{18}=0,1\)
\(n_{O_2}=\frac{3,36}{22,4}=0,15\)
Số mol O2 phản ứng ở phản ứng đầu là: \(0,15-0,05=0,1\)
\(\Rightarrow m_{CO_2}=0,2.44=8,8\)
b/ \(m_{CO}=0,2.28=5,6\)
\(m_{H_2}=0,1.2=0,2\)
c/ \(\%CO=\frac{0,2}{0,3}.100\%=66,67\%\)
\(\Rightarrow\%H_2=100\%-66,67\%=33,33\%\)
\(2CO + O_2 \xrightarrow{t^o} 2CO_2\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{CO} = n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{n_{CO} + n_{H_2}}{2}=\dfrac{0,2+n_{H_2}}{2} = \dfrac{9,6}{32} = 0,3(mol)\\ \Rightarrow n_{H_2} = 0,4(mol)\\ \%V_{CO} = \dfrac{0,2}{0,2 + 0,4}.100\% = 33,33\%\\ \%V_{H_2} = 100\% - 33,33\% = 66,67\%\\ \%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\%=87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)
Giả sử các khí được đo ở điều kiện sao cho 1 mol khí chiếm thể tích 1 lít
Gọi số mol CH4, C2H6 là a, b (mol)
=> \(a+b=\dfrac{25}{1}=25\left(mol\right)\) (1)
\(n_{O_2}=\dfrac{95}{1}=95\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
a---->2a---------->a
2C2H6 + 7O2 --to--> 4CO2 + 6H2O
b------>3,5b-------->2b
=> \(\left\{{}\begin{matrix}n_{O_2\left(dư\right)}=95-2a-3,5b\left(mol\right)\\n_{CO_2}=a+2b\left(mol\right)\end{matrix}\right.\)
=> \(95-a-1,5b=\dfrac{60}{1}=60\)
=> a + 1,5b = 35 (2)
(1)(2) => a = 5; b = 20
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{5}{25}.100\%=20\%\\\%V_{C_2H_6}=\dfrac{20}{25}.100\%=80\%\end{matrix}\right.\)
\(\overline{M}_A=\dfrac{5.16+20.30}{5+20}=27,2\left(g/mol\right)\)
\(\overline{M}_B=20,5.2=41\left(g/mol\right)\)
=> \(d_{A/B}=\dfrac{27,2}{41}\approx0,663\)
\(a)\\ 2CO + O_2 \xrightarrow{t^o} 2CO\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{H_2} = n_{H_2O} = \dfrac{1,8}{18} = 0,1(mol)\\ \)
Theo PTHH :
\(2n_{O_2} = n_{CO} + n_{H_2}\\ \Leftrightarrow 2.\dfrac{3,36}{22,4} = n_{CO} + 0,1\\ \Leftrightarrow n_{CO} = 0,2(mol)\\ \%V_{H_2} = \dfrac{0,1}{0,1+ 0,2}.100\% = 33,33\%\\ \%V_{CO} = 100\%-33,33\% = 66,67\%\\ c) Cách\ 1 :\\ n_{CO_2} = n_{CO} = 0,2(mol)\\ m_{CO_2} = 0,2.44 = 8,8(gam)\\ Cách\ 2 : \\ m_{hh} = m_{CO} + m_{H_2} = 0,2.28 + 0,1.2 = 5,8(gam) \)
Bảo toàn khối lượng :
\(m_{hh} + m_{O_2} = m_{H_2O} + m_{CO_2}\\ \Rightarrow m_{CO_2} = 5,8 + 0,15.32 - 1,8 = 8,8(gam)\)