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\(n_{CH_4}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,5 1 0,5 ( mol )
\(V_{O_2}=n.24,79=1.24,79=24,79l\)
\(V_{CO_2}=n.24,79=0,5.24,79=12,395l\)
nCH4 = 11,2/22,4 = 0,5 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,5 ---> 1
VO2 = 1 . 24,79 = 24,79 (l)
1)
$CH_4 +2 O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
Theo PTHH :
$V_{O_2\ cần\ dùng} = 2V_{CH_4} = 24,79(lít)$
$V_{CO_2} = V_{CH_4} = 12,395(lít)$
2)
a)
$C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$
$V_{O_2} = 3V_{C_2H_4} = 14,874(lít)$
b) $V_{không\ khí} = V_{O_2} : 20\% = 14,874 : 20\% = 74,37(lít)$
PTHH: 2C2H2 + 5O2 --to--> 4CO2 + 2H2O
\(\dfrac{V_{O_2}}{V_{C_2H_2}}=\dfrac{n_{O_2}}{n_{C_2H_2}}=\dfrac{5}{2}\Rightarrow V_{O_2}=2,5.\dfrac{5}{2}=6,25\left(l\right)\)
=> \(V_{kk}=6,25.5=31,25\left(l\right)\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{14,874}{22,79}=0,6\left(mol\right)\\ \Rightarrow n_{CO_2}=n_{CH_4}=0,6\left(mol\right)\\ n_{O_2}=n_{H_2O}=2.0,6=1,2\left(mol\right)\\ V_{O_2\left(đkc\right)}=1,2.24,79=29,748\left(l\right)\\ V_{kk\left(đkc\right)}=29,748.5=148,74\left(l\right)\\ V_{CO_2\left(đkc\right)}=0,6.24,79=14,874\left(l\right)\\ m_{CO_2}=44.0,6=26,4\left(g\right)\\ m_{H_2O}=1,2.18=21,6\left(g\right)\\ V_{H_2O}=\dfrac{21,6}{1}=21,6\left(ml\right)\)
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: \(n_{CO_2}=n_{CH_4}=\dfrac{7437}{11200}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=\dfrac{7437}{11200}.22,4=14,874\left(lít\right)\\m_{CO_2}=\dfrac{7437}{11200}.44\approx29,22\left(g\right)\end{matrix}\right.\)
a, nC2H4 = 2,479/24,79 = 0,1 (mol)
PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O
Mol: 0,1 ---> 0,3
VO2 = 0,3 . 24,79 = 7,437 (l)
b, PTHH: 2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O
Mol: 0,12 <--- 0,3
VC2H2 = 0,12 . 24,79 = 2,9748 (l)
a. \(n_{CH_4}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)
PTHH : CH4 + 2O2 ---t0---> CO2 + 2H2O
0,2 0,4 0,2
b. \(V_{O_2}=0,4.22,4=8,96\left(l\right)\)
\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
c. \(V_{kk}=8,96.5=44,8\left(l\right)\)
a, PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Ta có: \(n_{C_2H_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,75\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,75.2,24=16,8\left(l\right)\)
\(\Rightarrow V_{kk}=16,8.5=84\left(l\right)\)
b, Theo PT: \(n_{CO_2}=2n_{C_2H_4}=0,5\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{\downarrow}=m_{CaCO_3}=0,5.100=50\left(g\right)\)
\(m_{Ca\left(OH\right)_2}=0,5.74=37\left(g\right)\)
\(\Rightarrow m_{ddCa\left(OH\right)_2}=\dfrac{37.100}{2}=1850\left(g\right)\)
Bạn tham khảo nhé!
a, PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Ta có: \(n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,45\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,45.22,4=10,08\left(l\right)\)
b, Theo PT: \(n_{CO_2}=2n_{C_2H_4}=0,3\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,3.22,4=6,72\left(l\right)\)
c, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{\downarrow}=m_{CaCO_3}=0,3.100=30\left(g\right)\)
Bạn tham khảo nhé!
PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có \(n_{CH_4}=\dfrac{V_{CH_4}}{24,79}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
\(\Rightarrow n_{O_2}=1\left(mol\right)\)
\(\Rightarrow V_{O_2}=n_{O_2}.24,79=24,79\left(l\right)\)