tính khối lượng khí oxi cần dùng để đốt cháy hoàn toàn : c) hỗn hợp có 8g khí hiđro và 2g khí metan d) hỗn hợp có 0,15mol C và 0,125 mol S
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có PTHH
C + O2 \(\rightarrow\) CO2
mC(phản ứng) = 1 . 95% = 0.95 (tấn)
theo PT: 32g O2 : 12g C
theo ĐB : x tấn O2 : 0.95 tấn C
=> x = 0.95 x 32 : 12 =2.533 (tấn)
vậy mO2 = 2.533 tấn
b) ta có PTHH
CH4 + 2O2 \(\rightarrow\) 2H2O + CO2
Theo PT : 16g CH4 : 64g O2
Theo ĐB : 4kg CH4 : x kg O2
=> x = 4 x 64 : 16 =16 kg O2
c) Ta cóPTHH
2H2 + O2 \(\rightarrow\) 2H2O (1)
CH4 + 2O2 \(\rightarrow\) CO2 + 2H2O (2)
CÓ:
nH2 = m/M = 8/2 = 4 (mol)
nCH4 = m/M = 2/16 =0.125(mol)
theo PT(1)=> nO2 = 1/2 . nH2 = 1/2 . 4 =2 (mol)
theo PT(2) => nO2 =2 . nCH4 = 2 x 0.125 =0.25(mol)
=> tổng nO2 = 2 + 0.25 = 2.25(mol)
=> mO2 = n . M = 2.25 x 32 =72(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)
\(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,25\left(mol\right)\\n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,25.22,4}{6,72}.100\%\approx83,33\%\\\%V_{C_2H_2}\approx16,67\%\end{matrix}\right.\)
Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,625\left(mol\right)\Rightarrow m_{O_2}=0,625.32=20\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đây nha câu a
Ta có pt:
C+O2->CO2 (1)
S+O2->SO2 (2)
theo PT(1) => nO2 = nC = 0.15 (mol)
theo PT(2) => nO2 = nS = 0.125(mol)
=> tổng nO2 = 0.15 + 0.125 =0.275(mol)
=> mO2 = n . M = 0.275 x 32 =8.8(g)
1. PT:
C + O2 ---> CO2 (1)
S + O2 ---> SO2 (2)
Theo PT(1), ta có: nO = nC = 0,15(mol).
=> mO=0,15.32=4,8(g).
Theo PT(2), ta có: nO = nS = 0,125(mol).
=> mO=0,125.32=4(g).
2. PT:
2H2 + O2 ---> 2H2O (3)
CH4 + O2 ---> CO2 + 2H2O (4)
Ta có: PT(3), nH2=8/1=8(mol).
PT(4), nCH4=2/16=0,125(mol).
Theo PT(3), ta có: nO = 1/2 . 8=4(mol)
=> mO=4.32=128(g)
Theo PT(4), ta có: nO = nCH4=0,125(mol)
=> mO=0,125.32=4(g).
![](https://rs.olm.vn/images/avt/0.png?1311)
$n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)$
\(2CO+O_2\xrightarrow[]{t^o}2CO_2\)
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)
\(\%m_{CO}=\dfrac{0,2.44}{0,2.44+0,4.2}.100\%=91,67\%\\ \%m_{H_2}=100\%-91,67\%=8,33\%\)
\(\%n_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\ \%n_{H_2}=100\%-33,33\%=66,67\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C2H4+Br2->C2H4Br2
0,05----0,05
n Br2=\(\dfrac{8}{160}\)=0,05 mol
=>%VC2H4=\(\dfrac{0,05.22,4}{5,6}.100=20\%\)
=>%VCH4=80%
c)CH4+2O2-to>CO2+2H2O
1.10-3----2.10-3 mol
C2H4+3O2-to>2CO2+2H2O
2,5.10-4-7,5.10-4 mol
n hh=\(\dfrac{0,028}{22,4}\)=1,25.10-3 mol
=>n C2H4=2,5.10-4 mol
=>n CH4=1.10-3 mol
=>VO2=(2.10-3+7,5.10-4).22,4=0,0616l
\(n_{Br_2}=\dfrac{8}{160}=0,05mol\)
\(\Rightarrow n_{etilen}=n_{Br_2}=0,05mol\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(\Rightarrow n_{metan}=n_{hh}-n_{etilen}=0,25-0,05=0,2mol\)
a)\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b)\(\%V_{metan}=\dfrac{0,2}{0,25}\cdot100\%=80\%\)
\(\%V_{etilen}=100\%-80\%=20\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mS = 4 . 40% = 1,6 (g)
mC = 4 - 1,6 = 2,4 (g)
nS = 1,6/32 = 0,05 (mol)
nC = 2,4/12 = 0,2 (mol)
PTHH: S + O2 -> (t°) SO2
Mol: 0,05 ---> 0,05
C + O2 -> (t°) CO2
Mol: 0,2 ---> 0,2
nO2 (cần dùng) = 0,05 + 0,2 = 0,25 (mol)
VO2 (cần dùng) = 0,25 . 22,4 = 5,6 (l)
help mi
trl như qq