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![](https://rs.olm.vn/images/avt/0.png?1311)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{CO_2}=2.0,25=0,5\left(mol\right)\\ a,V_{CO_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{O_2}=\dfrac{5}{2}.0,25=0,625\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,625.22,4=14\left(l\right)\\ V_{kk\left(đkct\right)}=\dfrac{100}{20}.14=70\left(lít\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2C_2H_2+5O_2-^{t^o}\rightarrow4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{1,3}{26}=0,05\left(mol\right)\\ n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,125\left(mol\right)\\ \Rightarrow V_{O_2}=2,8\left(l\right)\\ Vìtrongkhôngkhíchứa20\%O_2\\ \Rightarrow V_{kk}=\dfrac{2,8}{20\%}=14\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PT: \(2C_2H_2+5O_2\underrightarrow{^{t^o}}4CO_2+2H_2O\)
\(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,5\left(mol\right)\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
b, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=56\left(l\right)\)
c, - Hiện tượng: Br2 nhạt màu dần.
PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nCH4 = 2.24/22.4 = 0.1 (mol)
CH4 + 2O2 -to-> CO2 + 2H2O
0.1____0.2______0.1
VO2 = 0.2*22.4 = 4.48 (l)
VCO2 = 0.1*22.4=2.24 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Theo PT: \(n_{C_2H_4}=\dfrac{1}{3}n_{O_2}=0,5\left(mol\right)\)
\(\Rightarrow V_{C_2H_4}=0,5.22,4=11,2\left(l\right)\)
Đáp án: C
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{C_2H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)
0,1 0,25 ( mol )
\(V_{kk}=0,25.22,4.5=28l\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,25-->0,5-------->0,25
=> \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
\(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,5--->1
=> VO2 = 1.22,4 = 22,4 (l)
b)
VO2 = 22,4.2 = 44,8 (l)
=> Vkk = 44,8.5 = 224 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{C_2H_4}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(C_2H_4+3O_2\underrightarrow{^{t^0}}2CO_2+2H_2O\)
\(0.2.........0.6........0.4..........0.4\)
\(V_{O_2}=0.6\cdot22.4=13.44\left(l\right)\)
\(m_{H_2O}=0.4\cdot18=7.2\left(g\right)\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
\(..............0.4.....0.4\)
\(m_{CaCO_3}=0.4\cdot100=40\left(g\right)\)
Ta có phương trình hóa học sau:+) \(2C_2H_2+5O_2\rightarrow4CO_2+2H_2O\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo phương trình hóa học trên, số mol C2H2 cần dùng là:
\(\dfrac{2\cdot0,3}{5}=0,12\left(mol\right)\)
➢\(V_{C_2H_2}=n_{C_2H_2}\cdot22,4=0,12\cdot22,4=2,688\left(l\right)\)
➤V\(C_2H_2\) = 2,688 lít