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nO2 = 9.6/32 = 0.3 (mol)
nCO2 = 8.8/44 = 0.2 (mol)
CO + 1/2O2 -to-> CO2
0.2_____0.1______0.2
H2 + 1/2O2 -to-> H2O
0.4__0.3-0.1
%CO = 0.2*28/(0.2*28 + 0.4*2) * 100% = 87.5%
Chúc bạn học tốt !!!
Mk củm chúc bn hc tốt , chậc bài nào bn cx giải giúp mìn hết lun á , khum bít ns j nên CỦM ƠN NHA ( Hc giỏi gơ á )
\(2CO + O_2 \xrightarrow{t^o} 2CO_2\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{CO} = n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{n_{CO} + n_{H_2}}{2}=\dfrac{0,2+n_{H_2}}{2} = \dfrac{9,6}{32} = 0,3(mol)\\ \Rightarrow n_{H_2} = 0,4(mol)\\ \%V_{CO} = \dfrac{0,2}{0,2 + 0,4}.100\% = 33,33\%\\ \%V_{H_2} = 100\% - 33,33\% = 66,67\%\\ \%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\%=87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)
nCO2 = 8.8/44 = 0.2 (mol)
nO2 = 9.6/32 = 0.3 (mol)
2CO + O2 -to-> 2CO2
0.2____0.1______0.2
2H2 + O2 -to-> 2H2O
0.4___0.3-0.1
%CO = 0.2*28 / ( 0.2*28 + 0.4*2) * 100% = 87.5%
%H2 = 12.5%
=> D
\(2CO + O_2\xrightarrow{t^o} 2CO_2(1)\\ n_{CO} = n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{n_{CO_2}}{2} = 0,1(mol)\\ 2H_2 +O_2 \xrightarrow{t^o} 2H_2O(2)\\ n_{H_2} = 2n_{O_2(2)} = 2.(\dfrac{9,6}{32}-0,1) = 0,4(mol)\\ \Rightarrow \%m_{CO} = \dfrac{0,2.28}{0,2.28 + 0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)
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: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)
PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\) (1)
\(2H_2+O_2\underrightarrow{t^o}2H_2O\) (2)
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_{CO}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow n_{O_2\left(1\right)}=0,1\left(mol\right)\\\Sigma n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(2\right)}=0,2\left(mol\right)\) \(\Rightarrow n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow\%V_{H_2}=\dfrac{0,4}{0,4+0,2}\cdot100\%\approx66,67\%\)
\(\Rightarrow\%V_{CO}=33,33\%\)
$n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)$
\(2CO+O_2\xrightarrow[]{t^o}2CO_2\)
0,2 0,1 0,2 (mol)
$n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)$
\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)
0,4 0,2 0,2 (mol)
\(\%m_{CO}=\dfrac{0,2.44}{0,2.44+0,4.2}.100\%=91,67\%\\ \%m_{H_2}=100\%-91,67\%=8,33\%\)
\(\%n_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\ \%n_{H_2}=100\%-33,33\%=66,67\%\)
Gọi \(\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO}=b\left(mol\right)\end{matrix}\right.\)⇒ 2a + 28b = 6,8(1)
\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ 2CO + O_2 \xrightarrow{t^o} 2CO_2\)
Theo PTHH :
\(n_{O_2} = 0,5a + 0,5b = \dfrac{8,96}{22,4} = 0,4(2)\)
Từ (1)(2) suy ra: a = 0,6 ; b = 0,2
Vậy :
\(\%m_{H_2} = \dfrac{0,6.2}{6,8}.100\% = 17,65\%\\ \%m_{CO} = 100\% - 17,65\% = 82,35\%\)
Cho em hỏi tại sao no2=0.5a+0.5b=0.4
tại sao viết 0.5 mà ko là 1 ạ
\(n_{CO_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH:
2CO + O2 --to--> 2CO2
0,2 0,1 0,2
-> nO2 = 0,3 - 0,1 = 0,2 (mol)
2H2 + O2 --to--> 2H2O
0,4 0,2
\(\rightarrow n_{hhkhí}=0,1+0,4=0,5\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,5}=40\%\\\%V_{H_2}=100\%-40\%=60\%\end{matrix}\right.\)
CO + \(\frac{1}{2}\)O2→CO2 H2 + \(\frac{1}{2}\)O2→H2O
6,6g CO2 :0,15mol → mol CO:0,15mol. →mol O2(khi p/ư CO) :0,075mol
Bài cho tổng mol O2:\(\frac{6,4}{32}\)=0,2mol →mol O2(khi p/ư H2 ) :0,2-0,075=0,125mol →mol H2:0,25mol
mkl=mco+mh2=0,15\(\times28+0,25\times2\)=4,7g →%mCO=\(\frac{0,15\times28}{4,7}\)=89,3%→%mH2=10,7%