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![](https://rs.olm.vn/images/avt/0.png?1311)
nCH4 = 6.72: 22,4 = 0.3 (mol)
a)
PTHH CH4 + 2O2 → CO2 + 2H2O
1 2 1 2 (mol)
0,15 0,3 0,15 0,3
b)
VO2 = n.22,4 = 0,6 . 22,4 = 13.44 (l)
⇒V kk = V O2 . 5 = 13.44 . 5 = 67.2 (l)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=n_{CH_4}=0,3\left(mol\right)\\n_{O_2}=n_{H_2O}=2n_{CH_4}=0,6\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{CO_2}=0,3.44=13,2\left(g\right)\)
\(m_{H_2O}=0,6.18=10,8\left(g\right)\)
\(V_{O_2}=0,6.22,4=13,44\left(l\right)\Rightarrow V_{kk}=13,44.5=67,2\left(l\right)\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_{CH_4}=\dfrac{6,4}{16}=0,4\left(mol\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
___0,4____0,8___0,4____0,8 (mol)
a, Ta có: \(m_{CO_2}=0,4.44=17,6\left(g\right)\)
\(m_{H_2O}=0,8.18=14,4\left(g\right)\)
b, \(V_{O_2}=0,8.22,4=17,92\left(l\right)\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{14,874}{22,79}=0,6\left(mol\right)\\ \Rightarrow n_{CO_2}=n_{CH_4}=0,6\left(mol\right)\\ n_{O_2}=n_{H_2O}=2.0,6=1,2\left(mol\right)\\ V_{O_2\left(đkc\right)}=1,2.24,79=29,748\left(l\right)\\ V_{kk\left(đkc\right)}=29,748.5=148,74\left(l\right)\\ V_{CO_2\left(đkc\right)}=0,6.24,79=14,874\left(l\right)\\ m_{CO_2}=44.0,6=26,4\left(g\right)\\ m_{H_2O}=1,2.18=21,6\left(g\right)\\ V_{H_2O}=\dfrac{21,6}{1}=21,6\left(ml\right)\)
Đề cho đkc nên anh tính theo đkc nhé!
\(pthh:CH_4+2O_2\overset{t^o}{--->}CO_2\uparrow+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{14,874}{22,4}=\dfrac{7437}{11200}\left(mol\right)\)
Theo pt: \(n_{O_2}=n_{H_2O}=2.n_{CH_4}=2.\dfrac{7437}{11200}\approx1,328\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=1,328.22,4=29,7472\left(lít\right)\\m_{H_2O}=1,328.18=23,904\left(g\right)\end{matrix}\right.\)
Theo pt: \(n_{CO_2}=n_{CH_4}=\dfrac{7437}{11200}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=\dfrac{7437}{11200}.22,4=14,874\left(lít\right)\\m_{CO_2}=\dfrac{7437}{11200}.44\approx29,22\left(g\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\)
b)
\(n_{CH_4} = \dfrac{6,72}{22,4} = 0,3(mol)\\ n_{CO_2} = n_{CH_4} = 0,3(mol)\\ n_{H_2O} = 2n_{CH_4} = 0,6(mol)\\ \Rightarrow m_{sản\ phẩm} = 0,3.44 + 0,6.18 = 24(gam)\)
c)
\(n_{O_2} = 2n_{CH_4} = 0,3.2 = 0,6(mol)\\ \Rightarrow V_{O_2} = 0,6.22,4= 13,44(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} = 13.44.5 = 67,2(lít)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{C_2H_4}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
PTHH: \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)
0,2---->0,6---->0,4---->0,4
\(\Rightarrow V_{O_2}=0,6.24,79=14,874\left(l\right)\)
b) \(V_{CO_2}=0,4.22,4=9,916\left(l\right)\)
c) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)
0,4------>0,4
\(\Rightarrow\left\{{}\begin{matrix}x=m_{CaCO_3}=0,4.100=40\left(g\right)\\y=m_{b\text{ình}.t\text{ăng}}=m_{CO_2}+m_{H_2O}=0,4.44+0,4.18=24,8\left(g\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{C_2H_4}=\dfrac{13,44}{22,4}=0,6mol\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
0,6 1,8 1,2 1,2
a)\(V_{O_2}=1,8\cdot22,4=40,32l\)
\(V_{kk}=5V_{O_2}=5\cdot40,32=201,6l\)
b)\(m_{CO_2}=1,2\cdot44=52,8g\)
\(m_{H_2O}=1,2\cdot18=21,6g\)
c)\(n_{NaOH}=0,3\cdot2=0,6mol\)
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
1,2 0,6 0 0
0,3 0,6 0,3 0,3
0,9 0 0,3 0,3
\(m_{muối}=0,3\cdot106=31,8g\)
\(m_{H_2O}=0,3\cdot18=5,4g\)
nC2H4 = 13,44/22,4 = 0,6 (mol)
PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O
Mol: 0,6 ---> 1,8 ---> 1,2 ---> 1,2
VO2 = 1,8 . 22,4 = 40,32 (l)
Vkk = 40,32 . 5 = 201,6 (l)
mCO2 = 1,2 . 44 = 52,8 (g)
mH2O = 1,2 . 18 = 21,6 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CH_4}=\dfrac{m_{CH_4}}{M_{CH_4}}=\dfrac{1,6}{16}=0,1mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,1 0,2 0,1 ( mol )
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
0,1 0,1 ( mol )
\(m_{CaCO_3}=n_{CaCO_3}.M_{CaCO_3}=0,1.100=10g\)
\(V_{O_2}=n_{O_2}.22,4=0,2.22,4=4,48l\)
\(V_{kk}=V_{O_2}.5=4,48.5=22,4l\)
nCH4 = 1,6/16 = 0,1 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,1 ---> 0,2 ---> 0,1
CO2 + Ca(OH)2 -> CaCO3 + H2O
Mol: 0,1 ---> 0,1 ---> 0,1
mCaCO3 = 0,1 . 100 = 10 (g)
Vkk = 0,2 . 5 . 22,4 = 22,4 (l)
a) \(n_{C_4H_{10}}=\dfrac{11,6.10^3}{58}=200\left(mol\right)\)
PTHH: 2C4H10 + 13O2 --to--> 8CO2 + 10H2O
200---->1300-------->800---->1000
=> \(V_{O_2}=1300.24,79=32227\left(l\right)\)
b) \(m_{CO_2}=800.44=35200\left(g\right)\)
\(m_{H_2O}=1000.18=18000\left(g\right)\)
=> Tổng khối lượng sản phẩm = 35200 + 18000 = 53200 (g)
\(n_{C_4H_{10}}=\dfrac{11,6}{58}=0,2mol\)
Có \(p=1bar\)=0.986923267 atm\(\approx1l\)
Thể tích khí Oxi cần tìm:
\(n=\dfrac{p\cdot V}{R\cdot T}\Rightarrow V=\dfrac{n\cdot R\cdot T}{p}\)
\(\Rightarrow V=\dfrac{0,2\cdot0,082\cdot\left(25+273\right)}{0,99}=4,8872l\)
\(C_4H_{10}+\dfrac{13}{2}O_2\underrightarrow{t^o}4CO_2+5H_2O\)
0,2 0,8 1
\(m_{CO_2}=0,8\cdot44=35,2g\)
\(m_{H_2O}=1\cdot18=18g\)