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PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
a, Theo PT: \(n_{O_2}=2n_{CH_4}=1\left(mol\right)\)
\(\Rightarrow V_{O_2}=1.22,4=22,4\left(l\right)\)
b, \(V_{kk}=\dfrac{22,4}{20\%}=112\left(l\right)\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\Rightarrow n_{CO_2}=n_{CH_4}=0,5\left(mol\right);n_{O_2}=2.n_{CH_4}=2.0,5=1\left(mol\right)\\ V_{O_2\left(đktc\right)}=n_{O_2}.22,4=1.22,4=22,4\left(l\right)\\ V_{CO_2\left(đktc\right)}=n_{CO_2}.22,4=0,5.22,4=11,2\left(l\right)\)
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. \(V_{kk}=8,96.5=44,8\left(l\right)\)
nCH4 = 4,48/22,4 = 0,2 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,4
Vkk = 0,2 . 5 . 22,4 = 44,8 (l)
mCO2 = 0,2 . 44 = 8,8 (g)
mH2O = 0,4 . 18 = 7,2 (g)
PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2O
Mol: 0,2 <--- 0,2 ---> 0,2
mCaCO3 = 0,2 . 100 = 20 (g)
\(V_{O_2}=\dfrac{336}{5}=67,2\left(ml\right)=0,0672\left(l\right)\\ n_{O_2}=\dfrac{0,0672}{22,4}=0,003\left(mol\right)\\ CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CO_2}=n_{CH_4}=\dfrac{0,003}{2}=0,0015\left(mol\right)\\ a,V_{CH_4\left(đktc\right)}=0,0015.22,4=0,0336\left(l\right)\\ b,V_{CO_2\left(đktc\right)}=V_{CH_4\left(đktc\right)}=0,0336\left(l\right)\)
nCH4 =11,2/22,4 = 0,5 (mol)
PTHH CH4 + 2O2 -to-> CO2 + 2H2O
...........0,5.........1.............0,5............1
Vkk= 5. VO2 = 5. 22,4 .1 = 112 l
\(n_{CH_4}\)\(=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
\(0,5\) \(1\) \(\left(mol\right)\)
\(V_{O_2}=1.22,4=22,4\left(l\right)\)
\(V_{kk}=22,4:\dfrac{1}{5}=112\left(l\right)\)